Generating function for Stirling numbers, induction 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?
 A: Ok, let's try using induction as requested (I have a feeling it would be easier to use generating functions). 
Assume (for $|z| <1$)
$$S_k(z)=\sum_{n=k}^{\infty} \left[\begin{array}{c}n\\k\end{array}\right]\frac{z^n}{n!}=\frac{\log^k[1/(1-z)]}{k!}.$$
We like to evaluate
$$S_{k+1}(z)=\sum_{n=k+1}^{\infty} \left[\begin{array}{c}n\\k+1\end{array}\right]\frac{z^n}{n!}.$$
Note the recurrence relation
$$\begin{bmatrix} n \\ k\end{bmatrix} = \begin{bmatrix} n+1 \\ k+1\end{bmatrix}
- n \begin{bmatrix} n \\ k+1\end{bmatrix}. $$
With that we have
$$\begin{align}
S_k(z) &= \sum_{n=k}^{\infty}\begin{bmatrix} n+1 \\ k+1\end{bmatrix} \frac{z^n}{n!}
- \sum_{n=k}^\infty n \begin{bmatrix} n \\ k+1\end{bmatrix} \frac{z^n}{n!}\\
&= \sum_{n=k+1}^\infty  \begin{bmatrix} n \\ k+1\end{bmatrix} \frac{1}{(n-1)!} (
z^{n-1} - z^n)  \\
&=  (1- z)\frac{d}{d z} S_{k+1}(z), 
\end{align}$$
where we used the fact that $\begin{bmatrix} k \\ k+1 \end{bmatrix} =0$.
Solving for $S(k+1)$, we have (employing the substitution $t=\log[1/(1-z')]$ and the trivial fact that $S_{k+1}(0) =0$)
$$\begin{array} \\
 S_{k+1}(z) &=  \int_0^z \!dz'\, \frac{S_k(z')}{1-z'}
 = \int_0^z\! dz'\frac{\log^k[1/(1-z')]}{k! (1-z')} \\
&= \int_0^{\log[1/(1-z)]} \!dt\, \frac{t^k}{k!} =  \frac{\log^{k+1} [1/(1-z)]}{(k+1)!}.\end{array} $$
