Prove the identity involving summation and Stirling numbers of the second kind Prove the identity
$$(e^z-1)^m=m!\sum_{n}^{}{n \brace m}\frac{z^n}{n!}$$
$n\brace m$ stands for Stirling numbers of the second kind. I'm not really sure if $z$ is some special number or just an ordinary variable. And I have no clue on how to start.
Thanks!
 A: Suppose you are allowed to use the OGF
$$F(z) = \sum_{n\ge m} {n\brace m} z^n
= \prod_{r=1}^m \frac{z}{1-rz} 
= z^m \prod_{r=1}^m \frac{1}{1-rz}.$$
which   seems  like  a   reasonable  assumption   because  it   has  a
combinatorial proof.

Start by doing a partial fraction decomposition of the product term.
We have with it being rational that
$$\mathrm{Res}\left(\prod_{r=1}^m \frac{1}{1-rz}; z=\frac{1}{q}\right)
= \prod_{r=1}^{q-1} \frac{1}{1-r/q}
\times \left(-\frac{1}{q}\right)
\times \prod_{r=q+1}^m \frac{1}{1-r/q}
\\ = \left(-\frac{1}{q}\right) 
\times \prod_{r=1}^{q-1} \frac{q}{q-r}
\times \prod_{r=q+1}^m \frac{q}{q-r}
= - \frac{1}{q} q^{m-1} \frac{(-1)^{m-q}}{(q-1)! \times (m-q)!}
\\ = - \frac{1}{m!} q^{m-1} (-1)^{m-q} {m\choose q}.$$
This gives for the OGF $F(z)$ that
$$F(z) = -\frac{1}{m!} z^m 
\sum_{q=1}^m \frac{1}{z-1/q} 
q^{m-1} (-1)^{m-q} {m\choose q}.$$
In particular,
$$[z^n] F(z) = {n\brace m}
= -\frac{1}{m!} [z^n] z^m 
\sum_{q=1}^m \frac{1}{z-1/q} 
q^{m-1} (-1)^{m-q} {m\choose q}
\\= \frac{1}{m!} [z^{n-m}]
\sum_{q=1}^m \frac{q}{1-zq}
q^{m-1} (-1)^{m-q} {m\choose q}
\\ = \frac{1}{m!} 
\sum_{q=1}^m 
q^m q^{n-m} (-1)^{m-q} {m\choose q}
 = \frac{1}{m!} 
\sum_{q=1}^m 
q^n (-1)^{m-q} {m\choose q}.$$
To conclude introduce the EGF 
$$Q(z) = \sum_{n\ge 1} {n\brace m} \frac{z^n}{n!}$$
which is
$$ \sum_{n\ge 1} \frac{z^n}{n!}
\frac{1}{m!} 
\sum_{q=1}^m 
q^n (-1)^{m-q} {m\choose q}.$$
Interchanging the two summations this becomes
$$ \frac{1}{m!} 
\sum_{q=1}^m (-1)^{m-q} {m\choose q}
\sum_{n\ge 1} \frac{z^n q^n}{n!}
= \frac{1}{m!} 
\sum_{q=1}^m (-1)^{m-q} {m\choose q} 
(\exp(qz)-1)$$
The term at $q=0$ is zero, so we get
$$\frac{1}{m!} 
\sum_{q=0}^m (-1)^{m-q} {m\choose q} 
(\exp(qz)-1)
= \frac{1}{m!} 
\sum_{q=0}^m (-1)^{m-q} {m\choose q} \exp(qz)
\\ = \frac{1}{m!} (\exp(z)-1)^m.$$
The combinatorial proof of the OGF including some rather pedestrian work by this user can be found at this MSE link.
