I found this equation involving Stirling numbers of the second kind on Math World: $$\sum\limits_{m=1}^n (-1)^m(m-1)!\,S(n,m)=0$$ for every integer $n \geq 2$. Here, $S(n, m)$ denotes the appropriate Stirling number of the second kind (i.e., the number of set partitions of $\left\{1,2,\ldots,n\right\}$ into $k$ parts).
However, I do not know why this is true. I am looking for a proof or an explanation of this equation.
Here, as another answer I would go with the following recursive formula for $S(n,m)$ if $1\leq m\leq n$ :
$$S(n,m)=mS(n-1,m)+S(n-1,m-1) $$
Now if I denote :
$$\varphi(n):=\sum_{m=1}^n(-1)^m(m-1)!S(n,m) $$
You have for $n\in\mathbb{N}^*$ :
$$\varphi(n+1):=\sum_{m=1}^{n+1}(-1)^m(m-1)!S(n+1,m) $$
$$\varphi(n+1)=\sum_{m=1}^{n+1}(-1)^m(m-1)!(mS(n,m)+S(n,m-1)) $$
$$\varphi(n+1)=\sum_{m=1}^{n+1}(-1)^mm!S(n,m)+\sum_{m=1}^{n+1}(-1)^m(m-1)!S(n,m-1) $$
$$\varphi(n+1)=\sum_{m=1}^{n+1}(-1)^mm!S(n,m)+\sum_{m=0}^{n}(-1)^{m+1}m!S(n,m) $$
$$\varphi(n+1)=\sum_{m=1}^{n+1}(-1)^mm!S(n,m)-\sum_{m=0}^{n}(-1)^{m}m!S(n,m) $$
$$\varphi(n+1)=(-1)^{n+1}(n+1)!S(n,n+1)-(-1)^{0}0!S(n,0) $$
Now $S(n,n+1)=0$ and $S(n,0)=1$ if $n=0$ and $0$ if $n>0$ so we have :
$$\varphi(1)=1\text{ and }\varphi(n)=0\text{ if } n>1 $$
I prefer the notation $n\brace m$ for your $S(n,m)$. One of the basic identities for Stirling numbers of the second kind is
$$x^n=\sum_m{n\brace m}(-1)^{n-m}x^{\overline m}\;,\tag{1}$$
where $x^{\overline m}=\prod_{k=0}^{m-1}(x+k)$ is a rising factorial.
It follows directly from the more familiar identity $$x^n=\sum_m{n\brace m}x^{\underline m}\;,$$ where $x^{\underline m}=\prod_{k=0}^{m-1}(x-k)$ is the falling factorial.
Set $x=0$ in $(1)$; then for $n\ge 1$ we have
$$0=\sum_m{n\brace m}(-1)^{n-m}(m-1)!=(-1)^n\sum_m{n\brace m}(-1)^m(m-1)!\;,$$
and hence
$$\sum_m{n\brace m}(-1)^m(m-1)!=0\;,$$
as desired.
We have the identity, for $n\geq 2$, $$\sum_{k=1}^{n}S\left(n,k\right)\left(k-1\right)!z^{k}=\left(-1\right)^{n}\textrm{Li}_{1-n}\left(1+1/z\right) $$ where $\textrm{Li}_{n}\left(z\right) $ is the polylogarithm function, then if we take $z=-1 $ we have $$\sum_{k=1}^{n}S\left(n,k\right)\left(k-1\right)!\left(-1\right)^{k}=\left(-1\right)^{n}\textrm{Li}_{1-n}\left(0\right)=0. $$
Suppose we are trying to show that $$\sum_{m=1}^{n} (-1)^m (m-1)! {n\brace m} = 0.$$
Recall the species for set partitions which is $$\mathfrak{P}(\mathcal{U} \mathfrak{P}_{\ge 1}(\mathcal{Z}))$$ which gives the generating function $$G(z, u) = \exp(u(\exp(z)-1))$$
and hence $${n\brace m} = n! [z^n] \frac{(\exp(z)-1)^m}{m!}.$$
Substitute this into the sum to obtain $$n! [z^n] \sum_{m=1}^{n} (-1)^m (m-1)! \frac{(\exp(z)-1)^m}{m!} = n! [z^n] \sum_{m=1}^{n} \frac{(-1)^m}{m} (\exp(z)-1)^m.$$
We may extend the sum to infinity beyond $n$ because $$(\exp(z)-1)^m$$ starts at $z^m$ including for $m>n.$ This yields $$n! [z^n] \sum_{m=1}^{\infty} \frac{(-1)^m}{m} (\exp(z)-1)^m \\ = n! [z^n] \log\frac{1}{1+\exp(z)-1} = n! [z^n] \log\exp(-z) = n! [z^n] (-z).$$
Therefore the sum is minus one when $n=1$ and zero otherwise.
Addendum Thu Oct 15 2015. The proof of the log-exp identity is at this MSE link.
