<s>[INST] What is $\sum_{n=1}^n(-1)^m\frac{m^{n-m+1}}{(n-m)!}$? So consider the sum
$$S=\sum_{m=1}^n(-1)^m\frac{m^{n-m-1}}{(n-m)!},$$
where $n$ is some fixed, positive integer. For specific values of $n$, this gives $-1$, $-\frac{1}{2}$, $\frac{1}{6}$, $\frac{1}{12}$, $-\frac{3}{40}$, $-\frac{1}{120}$, $\frac{31}{1008}$, $-\frac{29}{5040}$, $-\frac{7}{640}$, $\frac{2087}{362880}$,... for $n=1, 2, 3, 4, 5, 6, 7, 8, 9, 10,$ ...  Is there a way to evaluate this sum for general $n$?
 [/INST] The generating function for
$$S_n = \sum_{m=1}^n (-1)^m \frac{m^{n-m-1}}{(n-m)!}$$
may be proved as follows:
$$\sum_{m=0}^{n-1} (-1)^{n-m} \frac{(n-m)^{m-1}}{m!}
= \frac{(-1)^n}{n}
+ \sum_{m=1}^{n-1} (-1)^{n-m} \frac{(n-m)^{m-1}}{m!}
\\ = \frac{(-1)^n}{n}
+ \sum_{m=1}^{n-1} (-1)^{n-m} \frac{1}{m} [z^{m-1}]
\exp((n-m)z)
\\ = \frac{(-1)^n}{n}
+ \sum_{m=1}^{n-1} (-1)^{n-m} \frac{1}{n-m} [z^m]
\exp((n-m)z)
\\ = \frac{(-1)^n}{n}
+ \sum_{m=1}^{n-1} (-1)^{m} \frac{1}{m} [z^{n-m}]
\exp(mz)
\\ = \sum_{m=1}^{n} (-1)^{m} \frac{1}{m} [z^{n-m}]
\exp(mz)
\\ = [z^n] \sum_{m=1}^{n} (-1)^{m} \frac{1}{m} z^m
\exp(mz).$$
Now here the  coefficient extractor enforces the range and we get
$$[z^n] \sum_{m\ge 1} (-1)^{m} \frac{1}{m} z^m \exp(mz)
= [z^n] \log\frac{1}{1+z\exp(z)}.$$
</s>