TITLE: How would I go about proving this expression?
QUESTION [8 upvotes]: Expression:
$$
\sum_{c=1}^{n-1} \sum_{k=c+1}^{n} \sum_{j=0}^{min\{c,k-c\}} {c \choose j} \frac{(k-c)!}{(k-c-j)!} f^{k-c-j} g^{c-j} B_{n,k}^{(g \diamond f)^c}(x) =$$
$$ \sum_{w=1}^n \sum_{m_1=1}^w \sum_{m_2 = 1}^w \frac{m_1!m_2!}{(w-m_1)!(w-m_2)!(m_1+m_2-w)!}f^{w-m_1}g^{w-m_2} \sum_{l=1}^{n-1} {n \choose l} B_{n-l,m_1}^f(x)  B_{l,m_2}^g(x)
$$
It is known that:
$$
B_{a,b}^f(x) = B_{a,b}^g(x) = 0 \quad\quad\quad (b > a)
$$
$B_{n,k}^f(x)$ is the partial bell polynomial of the function $f(x)$ where:
$$
B_{n,k}^f(x) = B_{n,k}\left(f'(x),f''(x),\ldots,f^{(n-k+1)}(x)\right)
$$
For more information on Bell Polynomials click here, but please be aware that no information regarding Bell Polynomials is needed to complete the proof.
Note that:
$$
B_{n,k}^{(g \diamond f)^c}(x) = \sum_{l=1}^{n-1} {n \choose l} B_{n-l,c}^f(x)B_{l,k-c}^g(x)
$$
My Observations:
I have no formal education in techniques for proving these kinds of expressions, but I will share what I have observed:
By setting 
$$m_1 = c$$
$$m_2 = k-c$$
$$w = k -j$$
We can see (disregarding the summations) that the expressions are exactly the same. This is what I mean:
$$
\sum_{c=1}^{n-1} \sum_{k=c+1}^{n} \sum_j {c \choose j} \frac{(k-c)!}{(k-c-j)!} f^{k-c-j} g^{c-j} \Phi(n,k,c) =
$$
$$ \sum_{k-j=1}^n \sum_{c=1}^{k-j} \sum_{k-c = 1}^{k-j}\frac{c!(k-c)!}{j!(k-c-j)!(c-j)!}f^{k-c-j}g^{c-j} \sum_{l=1}^{n-1} {n \choose l} \rho(l,c,f) \rho(n-l,k-c,g)
$$
Where as stated previously, the summations do not even make sense, but the terms inside the summations are identical. Not sure if this helps but it is interesting and it is something I observed.
We can also see that in order for the terms to be non-zero the following inequality must be true:
$$
m_1+m_2 \ge w
$$
This is because of the coefficients:
$$
\frac{m_1!m_2!}{(w-m_1)!(w-m_2)!(m_1+m_2-w)!}
$$
Lets assume that $m_1+m_2 \le w$, then we have $m_1+m_2 - w = -I$ where $-I$ is a negative whole number, therefore the coefficient would be $0$ by taking a negative factorial of an integer:
$$
\frac{m_1!m_2!}{(w-m_1)!(w-m_2)!(-I)!} = \frac{m_1!m_2!}{(w-m_1)!(w-m_2)!} \frac{1}{\pm \infty} = 0
$$
Since we know $w \ge m_1 \ge 1$, $w \ge m_2 \ge 1$, and $n \ge w \ge 1$
What I am looking for: I am looking for a good place or technique to start or complete the proof. Feel free to present your own proof if you wish as well.

REPLY [8 votes]: Note: This is a partial answer with  the goal to present an equivalent identity with reduced complexity. In order to do so we analyse the LHS and RHS of OPs expression and rearrange and align step by step both sides which enables us to eliminate equally shaped outer sums.
We claim the following

In order to prove the validity of OPs expression, it is sufficient to show for
$1\leq l\leq n-1,\ 1\leq c\leq n-l$
  \begin{align*}
\sum_{k=1}^{l}&\sum_{j=0}^{\min\{c,k\}}\frac{(c)_j(k)_j}{j!}f^{k-j}g^{c-j}B_{l,k}^g(x)\\
&=\sum_{j=0}^{l}\sum_{k=\max\{1,j\}}^{\min\{j+c,l\}}\frac{(c)_{k-j}(k)_j}{j!}f^jg^{j+c-k}B_{l,k}^g(x)\tag{1}
\end{align*}

Observe,  that the  original expression contains four sums on each side. So, this expression has a reduced complexity. Nevertheless some further tricky rearrangements are necessary to prove (1). 

Step 1:  Adjustment of LHS and RHS
At first we  rearrange both sums of OPs expression without renaming indices. For a better comparison we also write $B_{n,k}^{(g\circ f)^c}$  fully expanded.
The   LHS can be written as 
  \begin{align*}
\sum_{c=1}^{n-1} &\sum_{k=c+1}^{n} \sum_{j=0}^{\min\{c,k-c\}} \binom{c}{j} \frac{(k-c)!}{(k-c-j)!} f^{k-c-j} g^{c-j} B_{n,k}^{(g \diamond f)^c}(x)\\
&= \sum_{l=1}^{n-1}\sum_{c=1}^{n-1}\sum_{k=c+1}^{n} \sum_{j=0}^{\min\{c,k-c\}}
\binom{n}{l}\frac{(c)_j(k-c)_j}{j!} f^{k-c-j} g^{c-j} B_{n-l,c}^{f}(x)B_{l,k-c}^{g}(x)\tag{2}\\
&= \sum_{l=1}^{n-1}\sum_{c=1}^{n-l}\sum_{k=c+1}^{c+l} \sum_{j=0}^{\min\{c,k-c\}}
\binom{n}{l}\frac{(c)_j(k-c)_j}{j!} f^{k-c-j} g^{c-j} B_{n-l,c}^{f}(x)B_{l,k-c}^{g}(x)\tag{3}\\
\end{align*}
The RHS can be written as
  \begin{align*}
\sum_{w=1}^n& \sum_{m_1=1}^w \sum_{m_2 = 1}^w \frac{m_1!m_2!}{(w-m_1)!(w-m_2)!(m_1+m_2-w)!}f^{w-m_1}g^{w-m_2}\\
&\qquad\cdot\sum_{l=1}^{n-1} \binom{n}{l} B_{n-l,m_1}^f(x)  B_{l,m_2}^g(x)\\
&=\sum_{l=1}^{n-1} \sum_{w=1}^n \sum_{m_1=1}^w \sum_{m_2 = 1}^w\binom{n}{l}\frac{m_1!m_2!}{(w-m_1)!(w-m_2)!(m_1+m_2-w)!}f^{w-m_1}g^{w-m_2}\\
&\qquad\cdot B_{n-l,m_1}^f(x)  B_{l,m_2}^g(x)\tag{4}
\end{align*}

Comment: 

In (2) we expand $B_{n,k}^{(g\circ f)^c}$ and write the corresponding sum with index $l$ at the leftmost position. We also write the binomial coefficient $\binom{c}{j}$ and the factorials using the Pochhammer symbol $(r)_s:=r(r-1)\cdots (r-s+1)$.
In (3) we set the upper index of $c$ to $n-l$ since the partial Bell polynomial $B_{p,q}^{f}(x)=0$ whenever $q>p$.                     According to the factor $B_{n-l,c}^{f}(x)$ we have no contribution if $c>n-l$. Similarly, since we have a factor $B_{l,k-c}^{g}(x)$ we also have no contribution if $k-c>l$ and we can set the upper bound of $k$ to $\min\{n,c+l\}$. This minimum is equal to $c+l$, since $1\leq c \leq n-l$.
In (4) we write the sum with index $l$ at the leftmost position.


Step 2:  Adjustment of  LHS
We transform the LHS of OPs expression starting from (2). 
\begin{align*}
\sum_{l=1}^{n-1}&\sum_{c=1}^{n-l}\sum_{k=c+1}^{c+l} \sum_{j=0}^{\min\{c,k-c\}}
\binom{n}{l}\binom{c}{j}\frac{(k-c)!}{(k-c-j)!} f^{k-c-j} g^{c-j} B_{n-l,c}^{f}(x)B_{l,k-c}^{g}(x)\\
&=\sum_{l=1}^{n-1}\sum_{c=1}^{n-l}\sum_{k=1}^{l} \sum_{j=0}^{\min\{c,k\}}
\binom{n}{l}\binom{c}{j}\frac{k!}{(k-j)!} f^{k-j} g^{c-j} B_{n-l,c}^{f}(x)B_{l,k}^{g}(x)\tag{5}\\
&=\sum_{l=1}^{n-1}\binom{n}{l} B_{n-l,c}^{f}(x)\sum_{c=1}^{n-l}\sum_{k=1}^{l} \sum_{j=0}^{\min\{c,k\}}\frac{(c)_j(k)_j}{j!} f^{k-j} g^{c-j}B_{l,k}^{g}(x)\tag{6}\\
\end{align*}

Comment:

In (5) we shift the index $k$ by $c$ and replace each occurrence of $k$ with  $k+c$.
In (6) we put the factors $\binom{n}{l}$ and $B_{n-l,c}^{f}(x)$ at the leftmost position and we rewrite the other binomial coefficient and factorials using the Pochhammer symbol.


Step 3:  Adjustment of  RHS
We transform the LHS of OPs expression starting from (3). 
\begin{align*}
\sum_{l=1}^{n-1} &\sum_{w=1}^n \sum_{m_1=1}^w \sum_{m_2 = 1}^w\binom{n}{l}\frac{m_1!m_2!}{(w-m_1)!(w-m_2)!(m_1+m_2-w)!}\\
&\qquad\cdot f^{w-m_1}g^{w-m_2} B_{n-l,m_1}^f(x)  B_{l,m_2}^g(x)\\
&=\sum_{l=1}^{n-1}\sum_{w=1}^n \sum_{c=1}^w \sum_{m_2 = 1}^w
\binom{n}{l}\frac{c!m_2!}{(w-c)!(w-m_2)!(c+m_2-w)!}\\
&\qquad\cdot f^{w-c}g^{w-m_2} B_{n-l,c}^f(x)  B_{l,m_2}^g(x)\tag{7}\\
&=\sum_{l=1}^{n-1} \sum_{c=1}^n\sum_{w=c}^n  \sum_{m_2 = 1}^w
\binom{n}{l}\frac{c!m_2!}{(w-c)!(w-m_2)!(c+m_2-w)!}\\
&\qquad\cdot f^{w-c}g^{w-m_2} B_{n-l,c}^f(x)  B_{l,m_2}^g(x)\tag{8}\\
&=\sum_{l=1}^{n-1} \sum_{c=1}^{n-l}\sum_{w=c}^{c+l}  \sum_{m_2 = 1}^{\min\{w,l\}}
\binom{n}{l}\frac{c!m_2!}{(w-c)!(w-m_2)!(c+m_2-w)!}\\
&\qquad\cdot f^{w-c}g^{w-m_2} B_{n-l,c}^f(x)  B_{l,m_2}^g(x)\tag{9}\\
&=\sum_{l=1}^{n-1} \sum_{c=1}^{n-l}\sum_{w=0}^{l}  \sum_{m_2 = 1}^{\min\{w+c,l\}}
\binom{n}{l}\frac{c!m_2!}{w!(w+c-m_2)!(m_2-w)!}\\
&\qquad\cdot f^{w}g^{w+c-m_2} B_{n-l,c}^f(x)  B_{l,m_2}^g(x)\tag{10}\\
&=\sum_{l=1}^{n-1} \sum_{c=1}^{n-l}\sum_{w=0}^{l}  \sum_{m_2 = \max\{1,w\}}^{\min\{w+c,l\}}
\binom{n}{l}\frac{c!m_2!}{w!(w+c-m_2)!(m_2-w)!}\\
&\qquad\cdot f^{w}g^{w+c-m_2} B_{n-l,c}^f(x)  B_{l,m_2}^g(x)\tag{11}\\
&=\sum_{l=1}^{n-1} \sum_{c=1}^{n-l}\sum_{j=0}^{l}  \sum_{k = \max\{1,j\}}^{\min\{j+c,l\}}
\binom{n}{l}\frac{c!k!}{j!(j+c-k)!(k-j)!}f^{j}g^{j+c-k} B_{n-l,c}^f(x)  B_{l,k}^g(x)\tag{12}\\
&=\sum_{l=1}^{n-1} \binom{n}{l}B_{n-l,c}^f(x)  \sum_{c=1}^{n-l}\sum_{j=0}^{l}  \sum_{k = \max\{1,j\}}^{\min\{j+c,l\}}\frac{(c)_{k-j}(k)_j}{j!}f^{j}g^{j+c-k} B_{l,k}^g(x)\tag{13}\\
\end{align*}

Comment:

In (7) we exchange the variable $m_2$ with $c$. 
In (8) we exchange the sums with index $c$ and index $w$ by appropiately resetting their upper indices.
In (9) we change the upper bound of the index $c$ to $n-l$ due to $B_{n-l,c}^f(x)$ and we change the upper bound of $m_2$ to $\min\{w,l\}$ due to $B_{l,m_2}^g(x)$.
In (10) we shift the index $w$ by $c$ and replace each occurrence of $w$ by $w+c$.
In (11) we change the lower bound of $m_2$ to $\max\{1,w\}$ since we have to respect the factor $(m_2-w)!$, so that $m_2-w\geq 0$.
In (12) we rename $w$ with $j$ and $m_2$ with $k$.
In (13) we put the factors $\binom{n}{l}$ and $B_{n-l,c}^{f}(x)$ at a leftmost position and we rewrite the other binomial coefficient and factorials using Pochhammer symbol


Now it's time to harvest:
Step 4: Putting all together
When looking at the expression (6) which is equivalent to the LHS of OPs identity and the expression (13) which is a equivalent to the RHS of OPs identity, we obtain
\begin{align*}
\sum_{l=1}^{n-1}&\binom{n}{l} B_{n-l,c}^{f}(x)\sum_{c=1}^{n-l}\sum_{k=1}^{l} \sum_{j=0}^{\min\{c,k\}}\frac{(c)_j(k)_j}{j!} f^{k-j} g^{c-j}B_{l,k}^{g}(x)\\
&=\sum_{l=1}^{n-1} \binom{n}{l}B_{n-l,c}^f(x)  \sum_{c=1}^{n-l}\sum_{j=0}^{l}  \sum_{k = \max\{1,j\}}^{\min\{j+c,l\}}\frac{(c)_{k-j}(k)_j}{j!}f^{j}g^{j+c-k} B_{l,k}^g(x)\tag{14}
\end{align*}

Observe, the outer sums with index $l$ and $c$ coincide as well as the factor $\binom{n}{l}$ and the partial Bell polynomial $B_{n-l,c}^{f}(x)$. Since the exponents of $f$ and $g$ as well as the index of $B_{l,k}^g(x)$ uniquely identify the summands we can skip the two outer sums and consider the regions specified by the outer indices instead.

We conclude the following is equivalent to (14):
For $1\leq l \leq n-1,\ 1\leq c \leq n-l$
  \begin{align*}
\sum_{k=1}^{l}& \sum_{j=0}^{\min\{c,k\}}\frac{(c)_j(k)_j}{j!} f^{k-j} g^{c-j}B_{l,k}^{g}(x)\\
&=\sum_{j=0}^{l}  \sum_{k = \max\{1,j\}}^{\min\{j+c,l\}}\frac{(c)_{k-j}(k)_j}{j!}f^{j}g^{j+c-k} B_{l,k}^g(x)\tag{15}
\end{align*}
  proving the validity of claim (1).

$$ $$

Step 5: How to proceed?
There are some tricky index transformations necessary in order to show the validity of the above expression. To get a better feeling for these transformations the following tables present the relevant values of the LHS and RHS for $n=3$.
Table 1: LHS of (15) for $n=3$
\begin{array}{cccccccccc}
n&l&c&k&j&\qquad\frac{(c)_j(k)_j}{j!}&f^{k-j}&g^{c-j}&B_{l,k}^g&\qquad \text{RHS}\\
\hline\\
3&1&1&1&0&\qquad1& f^1&g^1&B_{1,1}^g&\qquad2\\
&&&&1&\qquad1&f^0&g^0&B_{1,1}^g&\qquad1\\
&&2&1&0&\qquad1&f^1&g^2&B_{1,1}^g&\qquad4\\
&&&&1&\qquad2&f^0&g^0&B_{1,1}^g&\qquad3\\
&2&1&1&0&\qquad1&f^1&g^1&B_{2,1}^g&\qquad6\\
&&&&1&\qquad1&f^0&g^0&B_{2,1}^g&\qquad5\\
&&&2&0&\qquad1&f^2&g^1&B_{2,2}^g&\qquad8\\
&&&&1&\qquad2&f^1&g^0&B_{2,2}^g&\qquad7\\
\end{array}
Table 2: RHS of (15) for $n=3$
\begin{array}{cccccccccc}
n&l&c&k&j&\qquad\frac{(c)_{k-j}(k)_j}{j!}&f^j&g^{j+c-k}&B_{l,k}^g&\qquad\text{RHS}\\
\hline\\
3&1&1&1&0&\qquad1& f^0&g^0&B_{1,1}^g&\qquad1\\
&&&&1&\qquad1&f^1&g^1&B_{1,1}^g&\qquad2\\
&&2&1&0&\qquad2&f^0&g^1&B_{1,1}^g&\qquad3\\
&&&&1&\qquad1&f^1&g^2&B_{1,1}^g&\qquad4\\
&2&1&1&0&\qquad1&f^0&g^0&B_{2,1}^g&\qquad5\\
&&&&1&\qquad1&f^1&g^1&B_{2,1}^g&\qquad6\\
&&&2&1&\qquad2&f^1&g^0&B_{2,2}^g&\qquad7\\
&&&&2&\qquad1&f^2&g^1&B_{2,2}^g&\qquad8\\
\end{array}

Note, that lines of table 1 and 2 with the same entry in the column RHS coincide. From this information we can derive a mapping of the indices and this could give some assistance how further transformations should be performed.