Proof of Faà di Bruno's formula using a convolution identity for Bell polynomials? 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.
 A: 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.
A: It is not full answer for your question but the method of proof might be such way
Taylor expansion of $f(g(x+h))$ can be written as
$$
f(g(x+h))=f(g(x))+h\frac{d}{dx} \left( f(g(x)) \right)+\frac{h^2 }{2!}\frac{d^2}{dx^2} \left(f(g(x)) \right)+\frac{h^3 }{3!}\frac{d^3}{dx^3} \left( f(g(x) \right)+.... \tag 1
$$
$$
f(g(x+h))=f \left( g(x)+hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....\right)
$$
$$
f(g(x+h))=f(g(x))+(hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....)f'(g(x))+\frac{(hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....)^2}{2!}f''(g(x))+\frac{(hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....)^3}{3!}f'''(g(x))+.....
$$
$$
f(g(x+h))=\sum_{k=0}^\infty \frac{(hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....)^k}{k!}f^{(k)}(g(x))
$$
$$
\frac{d^n }{dh^{n}}f(g(x+h))=\frac{d^n }{dh^{n}}\left(\sum_{k=0}^\infty \frac{(hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....)^k}{k!}f^{(k)}(g(x))\right)
$$
$$
\frac{d^n }{dh^{n}}f(g(x+h))|_{h=0}=\frac{d^n }{dh^{n}}\left(\sum_{k=0}^\infty \frac{(hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....)^k}{k!}f^{(k)}(g(x))\right)|_{h=0}
$$
If we use Equation 1 and we can get 
$$
\frac{d^n }{dx^{n}}\left( f(g(x)) \right)=\sum_{k=0}^\infty \frac{d^n }{dh^{n}} \left((hg'(x)+\frac{h^2 }{2!}g''(x)+\frac{h^3 }{3!}g'''(x)+....)^k \right)|_{h=0} \frac{ f^{(k)}(g(x))}{k!}
$$
You need to use binominal expansion formula after this. 
