TITLE: Proof of Faà di Bruno's formula using a convolution identity for Bell polynomials?
QUESTION [5 upvotes]: I have noticed there is an identity for Bell polynomials that can apply of Faà di Bruno's formula. This is a convolution identity that states:
$$
(x \ast y)_n = \sum_{j=1}^{n-1} {n \choose j} x_j y_{n-j},
$$
which is applied to Bell polynomials which an example is shown on this wiki page. But I have yet to see a proof of this formula using this identity. Since Faà di Bruno's formula can be expressed as:
$$
\frac{d^n}{dx^n}[f(g(x))] = \sum_{k=1}^{n}f^{(k)}(g(x)) B_{n,k}(g'(x),g''(x),\dots,g^{(n-k+1)}(x)).
$$
Is there a proof of this formula using the convolution identity for Bell polynomials, or would this be unnecessarily rigorous? Either way it would be interesting to see.

REPLY [2 votes]: For a sequence $x=\left(x_{1},x_{2},...,x_{N}\right)$, define the convolution with:
$$(x\ast x)_{n}=\sum_{j=1}^{n-1}{n \choose j}x_{j}x_{n-j}.$$
Multiple application of this convolution yields:
$$\underbrace{(x\ast x\ast ...\ast x)}_{k\ factors}=\sum_{i_{1}=1}^{N}\sum_{i_{2}=1}^{i_{1}-1}\sum_{i_{3}=1}^{i_{2}-1}...\sum_{i_{k}=1}^{i_{k-1}-1}{i_{1}\choose i_{2}}{i_{2}\choose i_{3}}...{i_{k-1}\choose i_{k}}x_{i_{1}-i_{2}}x_{i_{2}-i_{3}}...x_{i_{k-1}-{i_k}}x_{i_{k}}$$
$$=k!\sum_{n=1}^{N}\sum_{\pi_{n,k}}{n!\over j_1!j_2!...j_{n-k+1}!}\left({x_1\over 1!}\right)^{j_1}\left({x_2\over 2!}\right)^{j_2}...\left({x_{n-k+1}\over(n-k+1)!}\right)^{j_{n-k+1}}$$
$$=k!\sum_{n=1}^{N}B_{n,k}(x_1,x_2,...,x_{n-k+1}),$$
$\pi_{n,k}=\pi_{n,k}(j_1,j_2,...,j_{n-k+1})$ are the partitions of integer $n$ into exactly $k$ parts.
This is the relation between Bell polynomials and your convolution. For proving Faà di Bruno's formula, it is not necessary to use the convolution.