Let $X = S^1 \sqcup S^1 \sqcup S^1 \sqcup \cdots \sqcup S^1$, where there are $m$ disjoint circles. Let $Y \subset X^n$ be the set of $n$ ordered points on $X$ such that there is at least one point on each component. Note that each connected component of $Y$ is isomorphic to $(S^1)^n$. Let the group $G = (S^1)^m$ act on $X$, where the $k$-th factor of $G$ rotates the $k$-th component of $X$. So $G$ acts on $Y$. Also, let $S_m$ act by permuting the factors. Let $Z = S_m\backslash Y/G$. Each component of $Y/G$ is isomorphic to $(S^1)^{n-m}$. The $S_m$ action permutes the components, so the quotient is the union of a smaller number of copies of $(S^1)^{n-m}$. So we see that $\chi(Z)=0$ for $n-m>0$.
I now present a regular CW structure on $Z$ so that the number of $k-m$ cells is $S(n,k) c(k,m)$, proving your identity. A given cell of $Z$ will correspond to those arrangements of points on $X$ which lie in a given cyclic order. So, to specify a cell of $Z$, we need to say (1) which points are equal to each other and (2) how to arrange those piles of points around circles. The number of ways to group $n$ points into $k$ equality classes is $S(n,k)$ and the number of ways to arrange those $k$ classes around $m$ circles is $c(k,m)$. We now need to see that this is a triangulation.
So, fix a partition of $[n]$ into $k$ blocks, and an arrangement of those blocks around $m$ circles. Let $U^{\circ}$ be the set of points in $Z$ with that configuration, and let $U$ be the closure of $U$. We need to see that $U^{\circ}$ is the interior of a ball and there is a continuous map from the closed ball to $Z$ extending the inclusion of $U^{\circ}$.
Choose coordinates on $U^{\circ}$ to be the angles between the blocks. Let the number of blocks on the $i$-th circle be $c_i$, so we'll write $\theta_1^i$, $\theta_2^i$, ..., $\theta_{c_i}^i$ for the coordinates coming from the $i$-th circle. So $\sum_{i=1}^m c_i = k$ and $\sum_{r=1}^{c_i} \theta_r^i = 2 \pi$ for each $i$.
Then
$$U^{\circ} = \prod_{i=1}^m {\Large \{} (\theta_1^i, \theta_2^i, \ldots, \theta_{c_i}^i) : \sum_r \theta_r^i = 2 \pi, \ \theta_r^i >0 {\Large \}}$$.
Clearly, $U^{\circ}$ is the interior of the ball
$$U := \prod_{i=1}^m {\Large \{} (\theta_1^i, \theta_2^i, \ldots, \theta_{c_i}^i) : \sum_r \theta_r^i = 2 \pi, \ \theta_r^i \geq 0 {\Large \}}$$
It is also easy to build a map $U \to Z$ extending the inclusion of $U^{\circ}$. It is not quite an injection: Whenever there is an $i$ such that every $\theta^i_r$ is either $0$ or $2 \pi$, the various points which are formed by moving the position of the $2 \pi$ will be identified.
Best Answer
The first formula in Section 24.1.4.I.B in Handbook of mathematical functions with formulas, graphs, and mathematical tables, Dover edition, 1965 (Library of Congress Catalog Card Number: 65.12253) by Abramowitz and Stegun is $$ x^n=\sum_{m=0}^n S(n,m) x(x-1)\cdots(x-m+1). \tag{1}\label{1} $$ Assuming here $n>1$ (so that $S(n,0)=0$), dividing both sides by $x$, and then letting $x=0$, we get the desired identity $$\sum_{m=1}^n S(n,m)(-1)^m(m-1)!=0.$$
One may note that Riordan (An Introduction to Combinatorial Analysis, Wiley, 1958, formula (35) on p. 33) uses identity \eqref{1} to define the Stirling numbers $S(n,m)$, then deriving their combinatorial definition (as the number of ways of partitioning a set of $n$ elements into $m$ nonempty sets) -- cf. formula (38) on p. 33 and the third display from the bottom on p. 91 of the book.