<s>[INST] Stirling numbers of first kind in a power series We need to prove that $\cfrac{(-\ln{(1-x)})^k}{k!}=\displaystyle\sum_{n=k}^\infty \begin{bmatrix} n\\  k \end{bmatrix}\cfrac{x^n}{n!}$. For this, I've done induction in $k$. The case $k=1$ is trivial. When I do the inductive step, I consider the next:
\begin{equation}
\begin{split}
\cfrac{(-\ln{(1-x)})^{k+1}}{(k+1)!}&=\left(\cfrac{(-\ln{(1-x)})^k}{k!}\right)\left(\cfrac{-\ln{(1-x)}}{k+1}\right)\\
&=\frac{1}{k+1}\left(\sum_{n=k}^\infty\begin{bmatrix} n\\  k \end{bmatrix}\frac{x^n}{n!}\right)\left(\sum_{n=1}^{\infty}\frac{x^n}{n}\right)\\
&=\frac{1}{k+1}\sum_{n=k+1}^\infty x^n\sum_{m=k}^{n-1} \begin{bmatrix} m\\  k \end{bmatrix}\frac{1}{(n-m)m!}.
\end{split}
\end{equation}
So, now my problem is showing that $\cfrac{n!}{k+1}\displaystyle\sum_{m=k}^{n-1}\begin{bmatrix} m\\  k \end{bmatrix}\frac{1}{(n-m)m!}=\begin{bmatrix} n\\  k+1 \end{bmatrix}$. On other hand, with the recurrence relation we can deduce the next:
\begin{equation}
\begin{split}
\begin{bmatrix} n\\  k+1 \end{bmatrix}&=\begin{bmatrix} n-1\\  k \end{bmatrix}+(n-1)\begin{bmatrix} n-1\\  k+1 \end{bmatrix}\\
&=\begin{bmatrix} n-1\\  k \end{bmatrix}+(n-1)\left(\begin{bmatrix} n-2\\  k \end{bmatrix}+(n-2)\begin{bmatrix} n-2\\  k+1 \end{bmatrix}\right)\\
&=\begin{bmatrix} n-1\\  k \end{bmatrix}+(n-1)\begin{bmatrix} n-2\\  k \end{bmatrix}+(n-1)(n-2)\begin{bmatrix} n-2\\  k+1 \end{bmatrix}\\
&\ \ \vdots\\
&=\frac{(n-1)!}{(n-1)!}\begin{bmatrix} n-1\\  k \end{bmatrix}+\frac{(n-1)!}{(n-2)!}\begin{bmatrix} n-2\\  k \end{bmatrix}+\frac{(n-1)!}{(n-3)!}\begin{bmatrix} n-3\\  k \end{bmatrix}+\ldots + \frac{(n-1)!}{k!}\begin{bmatrix} k\\  k \end{bmatrix}\\
&=(n-1)!\sum_{m=k}^{n-1}\frac{1}{m!}\begin{bmatrix} m\\  k \end{bmatrix}.
\end{split}
\end{equation}
 [/INST] We seek to prove that
$$\frac{1}{k!} \left(\log\frac{1}{1-z}\right)^k
= \sum_{n\ge k} {n\brack k} \frac{z^n}{n!}$$
by induction. We get for the base case
$$\log\frac{1}{1-z} = \sum_{n\ge 1} \frac{z^n}{n}$$
which holds by inspection. We also have
$$n! [z^n] \frac{1}{(k+1)!} \left(\log\frac{1}{1-z}\right)^{k+1}
\\ = (n-1)! [z^{n-1}]
\left(\frac{1}{(k+1)!} \left(\log\frac{1}{1-z}\right)^{k+1}\right)'
\\ = (n-1)! [z^{n-1}]
\left(\frac{1}{k!} \left(\log\frac{1}{1-z}\right)^{k}\right)
\left(\log\frac{1}{1-z}\right)'
\\ = (n-1)! [z^{n-1}]
\left(\frac{1}{k!} \left(\log\frac{1}{1-z}\right)^{k}\right)
\frac{1}{1-z}.$$
We now use the induction hypothesis and the fact that we have
a convolution of two EGFs to get for the coefficient being extracted
$$\sum_{q=k}^{n-1} {n-1\choose q} {q\brack k} (n-1-q)!$$
We may now conclude by combinatorics, re-writing the sum as follows:
$$\sum_{q=k}^{n-1} {n-1\choose n-1-q} (n-1-q)!  {q\brack k}$$
Here we are counting partitions of $[n]$ into $k+1$ cycles by classifying
according to the cycle where $n$ resides. We choose $n-1-q$ companions on
that  cycle where each choice generates $(n-q)!/(n-q)$ possible cycles.
The  remaining $q$ elements are partitioned into $k$ cycles. In this way
we  have counted all ${n\brack k+1}$ partitions exactly once, and we have
the claim.
 We may also continue algebraically using the OGF of the Stirling
numbers of the first kind:
$$\sum_{q=k}^{n-1} {n-1\choose n-1-q} (n-1-q)!
[w^k] {w+q-1\choose q} q!
\\ = (n-1)! [w^k] \sum_{q=k}^{n-1} {w+q-1\choose q}.$$
Now
$$\sum_{q\ge 0} {w+q-1\choose q} z^q = \frac{1}{(1-z)^w}$$
so we get
$$(n-1)! [w^k] [z^{n-1}] \frac{1}{(1-z)^{w+1}}
- (n-1)! [w^k] [z^{k-1}] \frac{1}{(1-z)^{w+1}}
\\ = (n-1)! [w^k] {w+n-1\choose n-1}
- (n-1)! [w^k] {w+k-1\choose k-1}
\\ = (n-1)! [w^k] \frac{n}{w} {w+n-1\choose n}
= n! [w^{k+1}] {w+n-1\choose n} = {n\brack k+1}.$$
Remark. In the induction step we have used the fact that when we
multiply two  exponential  generating functions  of the  sequences
$\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}.$$
</s>