I want to prove $\displaystyle\sum_{n=k}^{+\infty} \left[\begin{array}{c}n\\k\end{array}\right]\frac{z^n}{n!}=\frac{\log^k(1/(1-z))}{k!}$ by induction.
For $k=0$ or $k=1$ it works, suppose it works for some $k$, the question is if $\displaystyle\sum_{n=k+1}^{+\infty} \left[\begin{array}{c}n\\k+1\end{array}\right]\frac{z^n}{n!}=\frac{\log^{k+1}(1/(1-z))}{(k+1)!}$, so:
$\displaystyle\frac{\log^{k+1}(1/(1-z))}{(k+1)!}=\frac{1}{k+1}\cdot\frac{\log^k(1/(1-z))}{k!}\cdot\log(1/(1-z))= \\ \displaystyle\frac{1}{k+1}\cdot \left( \sum_{n=k}^{+\infty} \left[\begin{array}{c}n\\k\end{array}\right]\frac{z^n}{n!} \right)\cdot \left( \sum_{n=k}^{+\infty} (n-1)! \frac{z^n}{n!} \right)$ and if I'm not wrong we have convolution of exponential generating functions and it's equal to: $\displaystyle\frac{1}{k+1}\cdot \sum_n\sum_i{n \choose i}\left[\begin{array}{c}i\\k\end{array}\right](n-i+1)!\frac{z^n}{n!}$ but I don't know what to do next.. or maybe I went wrong way?