TITLE: Any ideas on how I can prove this expression?
QUESTION [27 upvotes]: I don't have a lot of places to turn because i am still in high school. So please bear with me as i had to create some notation. 
In order to understand my notation you must observe this identity for bell polynomials
$a = (f'(x),f''(x),\cdots)$ and $b = (g'(x),g''(x),\cdots)$
$$
B_{n,k}(f'(x),f''(x),\cdots,f^{(n-k+1)}(x))_{(f \rightarrow g)^c} = \frac{(a^{(k-c)_\diamond} \diamond b^{c_\diamond})_n}{(k-c)!c!}
$$
Also note that $d_n= \frac{d^n}{dx^n}[f(x)\ln(g(x))]$
I must prove that
$$
\sum_{k=1}^{n}\ln^k(g(x)) B_{n,k}(f'(x),f''(x),\cdots,f^{(n-k+1)}(x))
$$
$$
=\sum_{k=1}^n[ B_{n,k}(d_1,d_2,\cdots,d_{n-k+1})- \sum_{m=0}^{n-k}\sum_{j=0}^{m} {m \choose j} \frac{\ln^{m-j}(g(x))}{g(x)^k} \frac{d^j}{d(f(x))^j}[(f(x))_k] B_{n,m+k}(f'(x),\cdots,f^{(n-m-k+1)}(x))_{(f \rightarrow g)^k}]
$$
Where $(f(x))_k$ is the Pochhammer symbol for falling factorial
I have been trying to prove this for quite a while. Any advice on doing so would be amazing. Perhaps this can be put into a determinant or something of the sort, But I am not sure about that double summation. If you have advice PLEASE do so through a comment.

REPLY [7 votes]: Note: Since the body of an answer is limited to $30000$ characters, I had to split my answer. So, this is a follow-up containing the final part 3 of this answer
Note: [2015-03-08]
It's a great pleasure to provide now a complete answer to OPs challenging question. I could close the gap by adding Step 2 and Step 3 of Part 3 of my answer. It was a really tough job to perform all the calculations (and much more verification steps behind the scene). So, if the interested reader is willing to review parts of this answer any fruitful hints to reduce the size of this proof are appreciated.



Part 3:
Theme of this final part is to show that OPs expression in the form (12) is equal to the expression (38).
We show the validity of the following identity:
\begin{align*}
\sum_{k=1}^{n}&B_{n,k}^{f\cdot (\ln \circ g)}\\
&= \sum_{k=1}^{n}\left((\ln\circ g)^{k}B_{n,k}^f+\frac{(f)_k}{g^k}B_{n,k}^g\right)\tag{39}\\
&\qquad+\sum_{l=1}^{n-1}\sum_{m=1}^{l}\sum_{k=1}^{n-l}\sum_{j=0}^{m}\binom{n}{l}\binom{m}{j}
\frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{l,m}^{f}B_{n-l,k}^{g}\tag{40}\\
&=\sum_{k=1}^{n}\left((\ln \circ g)^k B_{n,k}^{f} + f^k  \sum_{j=k}^{n}\frac{(-1)^{j-k}}{g^j}\begin{bmatrix}j\\k\end{bmatrix}B^{g}_{n,j}\right)\tag{41}\\
&\qquad+\sum_{k=1}^{n}\sum_{m_1=1}^k\sum_{m_2=1}^{k}\frac{m_2!}{(k-m_1)!}\binom{m_1}{k-m_2}f^{k-m_1} (\ln \circ g)^{k-m_2}\\
&\qquad\qquad\cdot\sum_{l=1}^{n-1}\sum_{j=m_2}^{n-l}\frac{(-1)^{j-m_2}}{g^j}\begin{bmatrix}j\\m_2\end{bmatrix}\binom{n}{l}B_{l,m_1}^fB^{g}_{n-l,j}\tag{42}\\
\end{align*}

We will show the equality of this awesome looking identity in several steps. We start with the simpler parts (39) and (41). 

Part 3.1: Equality of (39) and (41)
The Stirling numbers of the first kind are often defined via the horizontal generating function $(x)_n$ of $(1+t)^x=\sum_{n\geq 0}(x)_n\frac{t^n}{n!}$, namely
\begin{align*}
(x)_n=\sum_{k=0}^{n}(-1)^{n-k}\begin{bmatrix}n\\k\end{bmatrix}x^{k}
\end{align*}
With this information at hand we can show the following identity is valid:
\begin{align*}
\sum_{k=1}^{n}\frac{(f)_k}{g^k}B_{n,k}^g
=\sum_{k=1}^{n}f^k  \sum_{j=k}^{n}\frac{(-1)^{j-k}}{g^j}\begin{bmatrix}j\\k\end{bmatrix}B^{g}_{n,j}\tag{43}
\end{align*}

We observe
\begin{align*}
\sum_{k=1}^{n}&f^k  \sum_{j=k}^{n}\frac{(-1)^{j-k}}{g^j}\begin{bmatrix}j\\k\end{bmatrix}B^{g}_{n,j}\\
&=\sum_{j=1}^{n}\frac{1}{g^{j}}B_{n,j}^{g}\sum_{k=1}^{j}(-1)^{j-k}\begin{bmatrix}j\\k\end{bmatrix}f^k\\
&=\sum_{j=1}^{n}\frac{(f)_j}{g^{j}}B_{n,j}^{g}
\end{align*}
and (43) follows.

We conclude: The easy part of OPs expression (39) and the derived formula (41) coincide.



Part 3.2: Equality of (40) and (42)
We will now analyse the complex sums (40) of OPs expression and compare with (42) of the derived one.
We have to prove that following is valid:
\begin{align*}
\sum_{l=1}^{n-1}&\sum_{m=1}^{l}\sum_{k=1}^{n-l}\sum_{j=0}^{m}\binom{n}{l}\binom{m}{j}
\frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{l,m}^{f}B_{n-l,k}^{g}\\
&=\sum_{k=1}^{n}\sum_{m_1=1}^k\sum_{m_2=1}^{k}\frac{m_2!}{(k-m_1)!}\binom{m_1}{k-m_2}f^{k-m_1} (\ln \circ g)^{k-m_2}\tag{44}\\
&\qquad\qquad\cdot\sum_{l=1}^{n-1}\sum_{j=m_2}^{n-l}\frac{(-1)^{j-m_2}}{g^j}\begin{bmatrix}j\\m_2\end{bmatrix}\binom{n}{l}B_{l,m_1}^fB^{g}_{n-l,j}\qquad\qquad n\geq 2
\end{align*}
  It is sufficient to consider $n\geq 2$ due to the range of the index $l$.

Idea: In order to show the equality of the identity (44) we will transform the RHS so that parts of the transformed expression can be easily compared with corresponding parts of OPs expression on the LHS. This way we will be able to split of sums of both expressions and so we can stepwise reduce the complexity of the identity which is to prove.

Step 1: Separation of the sums with index $l$ and $m$ (resp. $m_1$) in (42)
The following is valid for $n \geq 2$:
\begin{align*}
\sum_{k=1}^{n}&\sum_{m_1=1}^k\sum_{m_2=1}^{k}(-1)^{j-m_2}\frac{m_2!}{(k-m_1)!}\binom{m_1}{k-m_2}\begin{bmatrix}j\\m_2\end{bmatrix}f^{k-m_1}\\
&\qquad\qquad\cdot\frac{ (\ln \circ g)^{k-m_2}}{g^j}\sum_{l=1}^{n-1}\sum_{j=m_2}^{n-l}\binom{n}{l}B_{l,m_1}^fB^{g}_{n-l,j}\\
&=\sum_{l=1}^{n-1}\sum_{m_1=1}^{l}\binom{n}{l}B_{l,m_1}^f\sum_{k=m_1}^{n}\sum_{m_2=1}^{k}\sum_{j=m_2}^{n-1}(-1)^{j-m_2}\frac{m_2!}{(k-m_1)!}\\
&\qquad\qquad\cdot\binom{m_1}{k-m_2}\begin{bmatrix}j\\m_2\end{bmatrix}f^{k-m_1}
\frac{ (\ln \circ g)^{k-m_2}}{g^j}B^{g}_{n-l,j}\tag{45}
\end{align*}

The RHS of (45) is essentially a reshuffling of the LHS. We put the sum with index $l$ on the leftmost position and exchange the sums with index $k$ and $m_1$. Thereby we respect that the range $[1\ldots n]$ for $m_1$ is effectively  $[1\ldots l]$ since $B_{l,m_1}^f=0$ for $m_1>l$.
The benefit is, that the part 
$$\sum_{l=1}^{n-1}\sum_{m_1=1}^{l}\binom{n}{l}B_{l,m_1}^f$$
is also the first part of (40) in OPs expression. Therefore  we can split off this part and reduce the analysis to the rest of the expression. We claim

The following is valid for $n\geq 2$
\begin{align*}
\sum_{l=1}^{n-1}&\sum_{m=1}^{l}\binom{n}{l}B_{l,m}^{f}\sum_{k=1}^{n-l}\sum_{j=0}^{m}\binom{m}{j}
\frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{n-l,k}^{g}\\
&=\sum_{l=1}^{n-1}\sum_{m_1=1}^{l}\binom{n}{l}B_{l,m_1}^f\sum_{k=m_1}^{n}\sum_{m_2=1}^{k}\sum_{j=m_2}^{n-l}(-1)^{j-m_2}\frac{m_2!}{(k-m_1)!}\\
&\qquad\qquad\cdot\binom{m_1}{k-m_2}\begin{bmatrix}j\\m_2\end{bmatrix}f^{k-m_1}
\frac{ (\ln \circ g)^{k-m_2}}{g^j}B^{g}_{n-l,j}\tag{46}
\end{align*}

Since the first part $\sum_{l=1}^{n-1}\sum_{m_1=1}^{l}\binom{n}{l}B_{l,m_1}^f$ of the LHS and RHS of (46) coincides, we can split off this part and conclude:

It is sufficient to show for $n\geq 2, 1\leq l \leq n-1$ and $1\leq m,m_1\leq l$ the validity of
\begin{align*}
\sum_{k=1}^{n-l}&\sum_{j=0}^{m}\binom{m}{j}
\frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{n-l,k}^{g}\\
&=\sum_{k=m_1}^{n}\sum_{m_2=1}^{k}\sum_{j=m_2}^{n-l}(-1)^{j-m_2}\frac{m_2!}{(k-m_1)!}\\
&\qquad\qquad\cdot\binom{m_1}{k-m_2}\begin{bmatrix}j\\m_2\end{bmatrix}f^{k-m_1}
\frac{ (\ln \circ g)^{k-m_2}}{g^j}B^{g}_{n-l,j}\tag{47}
\end{align*}

In order to further simplify the identity, we take a look at the Bell polynomials of the LHS and RHS of (47)
$$B_{n-l,k}^{g} \longleftrightarrow B^{g}_{n-l,j}$$
Since they should coincide we rename the indices of the RHS of (47) in order to ease comparison with OPs expression.

Renaming of indices: We use following substitution
$$m_1 \longrightarrow m, \qquad m_2 \longrightarrow q, \qquad k \longleftrightarrow j$$
and obtain from (47)
The following is to prove for $n \geq 2, 1\leq l \leq n-1$ and $1\leq m \leq l$: 
\begin{align*}
\sum_{k=1}^{n-l}&\sum_{j=0}^{m}\binom{m}{j}
\frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{n-l,k}^{g}\\
&=\sum_{j=m}^{n}\sum_{q=1}^{j}\sum_{k=q}^{n-l}(-1)^{k-q}\frac{q!}{(j-m)!}\\
&\qquad\qquad\cdot\binom{m}{j-q}\begin{bmatrix}k\\q\end{bmatrix}f^{j-m}
\frac{ (\ln \circ g)^{j-q}}{g^k}B^{g}_{n-l,k}\tag{48}
\end{align*}

$$$$

Step 2: Separation of the sum with index $k$  in (48)
We observe, that the indices of the Bell polynomials of the LHS and RHS of (48) coincide. In order to further simplify the identity we transform the RHS so that the sum with index $k$ is at the leftmost position.

RHS:
\begin{align*}
\sum_{j=m}^{n}\sum_{q=1}^{j}\sum_{k=q}^{n-l}\ldots&=\sum_{q=1}^n\sum_{j=\max\{q,m\}}^n\sum_{k=q}^{n-l}\ldots
=\sum_{k=1}^{n-l}\sum_{q=1}^{k}\sum_{j=\max\{q,m\}}^n\ldots
\end{align*}

Applying this transformation we obtain
\begin{align*}
\sum_{k=1}^{n-l}&\frac{B_{n-l,k}^g}{g^k}\sum_{j=0}^m\binom{m}{j}\left(\ln\circ g\right)^{m-j}\frac{d^j}{d(f)^j}[(f)_k]\tag{49}\\
&=\sum_{k=1}^{n-l}\frac{B_{n-l,k}^g}{g^k}\sum_{q=1}^{k}\sum_{j=\max\{q,m\}}^n(-1)^{k-q}\frac{q!}{(j-m)!}
\binom{m}{j-q}\begin{bmatrix}k\\q\end{bmatrix}f^{j-m}(\ln \circ g)^{j-q}
\end{align*}

Comparing LHS and RHS of (49) we see, that we can split of the leftmost sum and it's sufficient to show the following:

The following is left to prove: For $n\geq 2, 1 \leq l \leq n-1, 1\leq k \leq n-l$ and $1 \leq m \leq l$
\begin{align*}
\sum_{j=0}^m&\binom{m}{j}\left(\ln\circ g\right)^{m-j}\frac{d^j}{d(f)^j}[(f)_k]\tag{50}\\
&=\sum_{q=1}^{k}\sum_{j=\max\{q,m\}}^n(-1)^{k-q}\frac{q!}{(j-m)!}
\binom{m}{j-q}\begin{bmatrix}k\\q\end{bmatrix}f^{j-m}(\ln \circ g)^{j-q}
\end{align*}

Next we will introduce the Stirling numbers of the first kind in the LHS of (50). According to (43) we know
$$(f)_k=\sum_{q=0}^{k}(-1)^{k-q}\begin{bmatrix}k\\q\end{bmatrix}f^q\qquad\qquad k\geq 1$$
It follows
\begin{align*}
\frac{d^j}{d(f)^j}[(f)_k]&=\frac{d^j}{d(f)^j}\sum_{q=0}^{k}(-1)^{k-q}\begin{bmatrix}k\\q\end{bmatrix}f^q\\
&=\sum_{q=j}^k(-1)^{k-q}\begin{bmatrix}k\\q\end{bmatrix}\frac{q!}{(q-j)!}f^{q-j}\qquad\qquad k\geq 1,j\geq 0
\end{align*}

Substituting this expression in (50) we obtain for $n\geq 2, 1\leq l \leq n-1, 1\leq k \leq n-l$ and $1\leq m \leq l$
  \begin{align*}
\sum_{j=0}^m&\binom{m}{j}\left(\ln \circ g\right)^{m-j}\sum_{q=j}^k(-1)^{k-q}
\begin{bmatrix}k\\q\end{bmatrix}\frac{q!}{(q-j)!}f^{q-j}\tag{51}\\
&=\sum_{q=1}^{k}\sum_{j=\max\{q,m\}}^n(-1)^{k-q}\frac{q!}{(j-m)!}
\binom{m}{j-q}\begin{bmatrix}k\\q\end{bmatrix}f^{j-m}(\ln \circ g)^{j-q}
\end{align*}

$$$$

Step 3: Separation of the sum with index $q$  in (51)
We get after exchanging the sums in the LHS of (51)
\begin{align*}
\sum_{q=1}^k&(-1)^{k-q}\begin{bmatrix}k\\q\end{bmatrix}q!
\sum_{j=0}^q\binom{m}{j}\left(\ln \circ g\right)^{m-j}\frac{1}{(q-j)!}f^{q-j}\\
&=\sum_{q=1}^{k}(-1)^{k-q}\begin{bmatrix}k\\q\end{bmatrix}q!
\sum_{j=\max\{q,m\}}^n\binom{m}{j-q}(\ln \circ g)^{j-q}\frac{1}{(j-m)!}f^{j-m}
\end{align*}
Observe that the index $q$ in the LHS now starts from $1$, since the Stirling Numbers of the first kind are always zero for $\begin{bmatrix}k\\0\end{bmatrix}$ with $k\geq 1$.
The following is left to prove for $n\geq 2, 1\leq l \leq n-1, 1\leq k \leq n-l,1\leq m \leq l$ and $1\leq q \leq k$
\begin{align*}
\sum_{j=0}^q&\binom{m}{j}\left(\ln \circ g\right)^{m-j}\frac{1}{(q-j)!}f^{q-j}\tag{52}\\
&=\sum_{j=\max\{q,m\}}^n\binom{m}{j-q}(\ln \circ g)^{j-q}\frac{1}{(j-m)!}f^{j-m}
\end{align*}

Short break for a moment: Comparing this expression (52) with the monstrous expression (39)-(42) from where we started, we can enjoy the considerable simplification. Yet (52) is tricky and not so easy to overcome due to reverse running indices $1\leq m \leq l$ and $1\leq k \leq n-l$ and $j$ on LHS with ($0\leq j\leq m$) and RHS ($\max\{q,m\}\leq j \leq n$) which walk to opposite ends of the region and some other (small) pitfalls. In order to proceed we will distinguish two cases.

Step 3.1: The case $q\leq m$
The following is left to prove for $n\geq 2, 1\leq l \leq n-1, 1\leq k \leq n-l,1\leq m \leq l$ and $1\leq q \leq k$ with the specific constraint $q\leq m$:
\begin{align*}
\sum_{j=0}^q&\binom{m}{j}\left(\ln \circ g\right)^{m-j}\frac{1}{(q-j)!}f^{q-j}\tag{52}\\
&=\sum_{j=m}^n\binom{m}{j-q}(\ln \circ g)^{j-q}\frac{1}{(j-m)!}f^{j-m}
\end{align*}
Now substituting $j \to q-j$ on the LHS of (52) gives:
LHS:
  \begin{align*}
\sum_{j=0}^q&\binom{m}{q-j}\left(\ln \circ g\right)^{m-q+j}\frac{1}{j!}f^{j}
\end{align*}
and shifting  $j \to j-m$ on the RHS gives:
  \begin{align*}
\sum_{j=0}^{n-m}&\binom{m}{j+m-q}(\ln \circ g)^{m-q+j}\frac{1}{j!}f^{j}\\
&=\sum_{j=0}^{n-m}\binom{m}{q-j}(\ln \circ g)^{m-q+j}\frac{1}{j!}f^{j}\\
&=\sum_{j=0}^{q}\binom{m}{q-j}(\ln \circ g)^{m-q+j}\frac{1}{j!}f^{j}\tag{53}
\end{align*}
and the claim (52) follows.

In (53) we changed the upper limit of the index $j$ by considering the following relationship between the indices:
Given: $n\geq 2, 1\leq l \leq n-1, 1\leq k \leq n-l,1\leq m \leq l, 1\leq q\leq k$ and the specific constraint $k\leq m$ we observe
\begin{equation*}
\left.
\begin{array}{l}
1\leq k \leq n-l\\
1\leq m\leq l
\end{array} \right\}
2\leq k+m\leq n\quad
\left.
\begin{array}{l}
k+m\leq n\\
q\leq k
\end{array} \right\}
q+m\leq n\quad\Rightarrow\quad q\leq n-m
\end{equation*}

Step 3.2: The case $m\leq q$
The following is left to prove for $n\geq 2, 1\leq l \leq n-1, 1\leq k \leq n-l,1\leq m \leq l$ and $1\leq q \leq k$ with the specific constraint $m\leq q$:
\begin{align*}
\sum_{j=0}^m&\binom{m}{j}\left(\ln \circ g\right)^{m-j}\frac{1}{(q-j)!}f^{q-j}\tag{54}\\
&=\sum_{j=q}^n\binom{m}{j-q}(\ln \circ g)^{j-q}\frac{1}{(j-m)!}f^{j-m}
\end{align*}
In the LHS of (54) the upper limit of the index $j$ changed from $q$ to $m$ since values greater than $m$ will nothing contribute due to $\binom{m}{j}$.
Now substituting $j \to m-j$ on the LHS of (54) gives:
LHS:
  \begin{align*}
\sum_{j=0}^m&\binom{m}{j}\left(\ln \circ g\right)^{m-q+j}\frac{1}{(j+q-m)!}f^{j+q-m}
\end{align*}
and shifting  $j \to j-q$ on the RHS gives:
  \begin{align*}
\sum_{j=0}^{n-q}&\binom{m}{j}(\ln \circ g)^{j}\frac{1}{j!}f^{j}\\
&=\sum_{j=0}^{n-q}\binom{m}{q-j}(\ln \circ g)^{m-q+j}\frac{1}{(j+q-m)!}f^{j+q-m}\\
&=\sum_{j=0}^{m}\binom{m}{q-j}(\ln \circ g)^{m-q+j}\frac{1}{(j+q-m)!}f^{j+q-m}\tag{55}
\end{align*}
and the claim (54) follows.

In (55) we changed the upper limit of the index $j$ by considering the following relationship between the indices:
Given: $n\geq 2, 1\leq l \leq n-1, 1\leq k \leq n-l,1\leq m \leq l, 1\leq q\leq k$ and the specific constraint $k\leq m$ we observe
\begin{equation*}
\left.
\begin{array}{l}
1\leq k \leq n-l\\
1\leq m\leq l
\end{array} \right\}
2\leq k+m\leq n\quad
\left.
\begin{array}{l}
k+m\leq n\\
q\leq k
\end{array} \right\}
q+m\leq n\quad\Rightarrow\quad m\leq n-q
\end{equation*}


Final Note: Part 3 of the proof demonstrates that OPs expression (39)+(40) is equal to the expression (41)+(42) which was created from scratch and so OPs expression is valid.
\begin{align*}
\sum_{k=1}^{n}&B_{n,k}^{f\cdot (\ln \circ g)}\\
&= \sum_{k=1}^{n}\left((\ln\circ g)^{k}B_{n,k}^f+\frac{(f)_k}{g^k}B_{n,k}^g\right)\tag{39}\\
&\qquad+\sum_{l=1}^{n-1}\sum_{m=1}^{l}\sum_{k=1}^{n-l}\sum_{j=0}^{m}\binom{n}{l}\binom{m}{j}
\frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{l,m}^{f}B_{n-l,k}^{g}\tag{40}\\
&=\sum_{k=1}^{n}\left((\ln \circ g)^k B_{n,k}^{f} + f^k  \sum_{j=k}^{n}\frac{(-1)^{j-k}}{g^j}\begin{bmatrix}j\\k\end{bmatrix}B^{g}_{n,j}\right)\tag{41}\\
&\qquad+\sum_{k=1}^{n}\sum_{m_1=1}^k\sum_{m_2=1}^{k}\frac{m_2!}{(k-m_1)!}\binom{m_1}{k-m_2}f^{k-m_1} (\ln \circ g)^{k-m_2}\\
&\qquad\qquad\cdot\sum_{l=1}^{n-1}\sum_{j=m_2}^{n-l}\frac{(-1)^{j-m_2}}{g^j}\begin{bmatrix}j\\m_2\end{bmatrix}\binom{n}{l}B_{l,m_1}^fB^{g}_{n-l,j}\tag{42}\\
\end{align*}


Note: This answer is based upon the following papers

V.V. Kruchinin_1: Composita and its properties
V.V. Kruchinin_2: Composition of ordinary generating functions
V.V. Kruchinin_3: Derivation of Bell Polynomials of the Second Kind
W.P. Johnson: The Curious History of Faa di Bruno's Formula


Note: Before I could finish the answer I've added this question to ask for support in solving the final gap.