Proof strategy - exponential generating functions Can anyone suggest how to touch this?
My task:

Prove that if sequences $a, b$ satisfy:
  $$
b_{n} = \sum_{k}^{}\left[n\atop k\right]a_{k}
$$
  then this equation is correct for their exponential generating functions $A, B$:
  $$

B(x) = A\left(\log\frac{1}{1-x}\right)

$$

Where $\left[n\atop k\right]$ are Stirling numbers of the first kind.
I would be grateful for any help.
 A: The generating function of Stirling numbers of the first kind is
$$
\sum_{k=0}^{+\infty} u^k \sum_{n=k}^{+\infty} \frac {x^n}{n!} \left[{n\atop k}\right] = \mathrm e^{ut},\qquad\text{with}\ t=\log(1/(1-x)).
$$
Expanding the exponential along the powers of $u$, this yields, for every $k\geqslant0$,
$$
\sum_{n=k}^{+\infty} \frac {x^n}{n!} \left[{n\atop k}\right] = \frac{t^k}{k!}.
$$
Hence,
$B(x)=\sum\limits_{n=0}^{+\infty} b_n\frac {x^n}{n!}$ is
$$
B(x)=\sum_{n=0}^{+\infty}\sum_{k=0}^n\left[{n\atop k}\right]a_k\frac {x^n}{n!}=\sum_{k=0}^{+\infty}a_k\sum_{n=k}^{+\infty}\left[{n\atop k}\right]\frac {x^n}{n!}=\sum_{k=0}^{+\infty}a_k\frac{t^k}{k!}=A(t).
$$
A: Let $A(x)$ be the exponential generating function for $a_n$, i.e. $\frac{d^n}{dx^n}A(0)=a_n$, and let $B(x)=A(\log(\frac{1}{1-x}))$.
We need to show that $\frac{d^n}{dx^n}B(0)=b_n$.
Let us find an expression for $\frac{d^n}{dx^n}B(x)$ using Stirling coefficients: $$\frac{d^n}{dx^n}B(x)=\sum_{k=1}^n\left[\matrix{n\\k}\right]a_k^n(x)A^{(k)}(-\log(1-x)).$$
We have $$\frac{d^{n+1}}{dx^{n+1}}B(x)=\sum_{k=1}^n\left[\matrix{n \\ k}\right]\left(a_k^{n'}(x)A^{(k)}(-\log(1-x))+\frac{a_k^n(x)}{1-x}A^{(k+1)}(-\log(1-x))\right)=a_1^{n'}(x)A^{(1)}(-\log(1-x))+\frac{a_n^n(x)}{1-x}A^{(n+1)}(-\log(1-x))+\sum_{k=2}^n\left(\left[\matrix{n \\ k}\right]a_k^{n'}(x)+\left[\matrix{n \\ k-1}\right]\frac{a_{k-1}^n(x)}{1-x}\right)A^{(k)}(-\log(1-x))=\sum_{k=1}^{n+1}\left(\left[\matrix{n \\ k}\right]a_k^{n'}(x)+\left[\matrix{n \\ k-1}\right]\frac{a_{k-1}^n(x)}{1-x}\right)A^{(k)}(-\log(1-x))$$
So we need $a_k^n(x)$ to be such that $\left[\matrix{n \\ k}\right]a_k^{n'}(x)+\left[\matrix{n \\ k-1}\right]\frac{a_{k-1}^n(x)}{1-x}=\left[\matrix{n+1 \\ k}\right]a_k^{n+1}(x)$. Also, we need $a_1^1(x)=\frac{1}{1-x}$.
Note that $n\left[\matrix{n \\ k}\right]+\left[\matrix{n \\ k-1}\right]=\left[\matrix{n+1 \\ k}\right]$. Therefore, we obtain the following expressions: $$a_k^n(x)=(1-x)^{-n}$$ and $$\frac{d^n}{dx^n}B(x)=\frac{1}{(1-x)^n}\sum_{k=1}^n\left[\matrix{n\\k}\right]A^{(k)}(-\log(1-x)).$$
Now just plug in $x=0$ into the final expression.
