How to prove $ \frac{\ln^k(1+x)}{k!}=\sum_{n=k}^\infty(-1)^{n-k} \begin{bmatrix} n \\ k \end{bmatrix}\frac{x^n}{n!}$ 
Prove the following formula involving Stirling numbers of the first kind:
  $$\frac{\ln^k(1+x)}{k!}=\sum_{n=k}^\infty(-1)^{n-k} \begin{bmatrix} n \\  k \end{bmatrix}\frac{x^n}{n!}$$
  where $\begin{bmatrix} n \\  k \end{bmatrix}$ is the Stirling number of the first kind.

I use this formula (can be found here) a lot in my solutions, but I have not found any proof of it yet. Any idea on how to prove it or where to find the proof?
I am tagging harmonic numbers as its very related to this formula.
Thank you. 

As said within the comment section it is sufficient to visit the Wikipedia Page and therefore no need to invoke some kind of Harmonic Numbers here. Within the subsection Generating Functions we eventually find the following paragraph:

A variety of identities may be derived by maniplulating the generating function:
  \begin{align*}
H(z,u)=(1+z)^u&=\sum_{n=0}^\infty\binom unz^n\\
&=\sum_{n=0}^\infty\frac{z^n}{n!}\sum_{k=0}^ns(n,k)u^k\\
&=\sum_{k=0}^\infty u^k\sum_{n=k}^\infty\frac{z^n}{n!}s(n,k)
\end{align*}
  Using the equality
  $$(1+z)^u=e^{u\log(1+z)}=\sum_{k=0}^\infty(\log(1+z))^k\frac{u^k}{k!}$$
  it follows that
  $$\sum_{n=k}^\infty(-1)^{n-k}\begin{bmatrix}n\\k\end{bmatrix}\frac{z^n}{n!}=\frac{(\log(1+z))^k}{k!}$$

The crucial relations used here are

\begin{align*}
&1.&&(x)_n~=~\sum_{k=0}^n s(n,k)x^k\\
&2.&&s(n,k)~=~(-1)^{n-k}\begin{bmatrix}n\\k\end{bmatrix}
\end{align*}

Which are, as far as I can tell (not being that experienced with Stirling Numbers at all), quite fundamental properties of the Stirling Numbers of the First Kind.
