What is $\sum_{n=1}^n(-1)^m\frac{m^{n-m+1}}{(n-m)!}$?

Question

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 $n$?

Answer 1

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

$$[z^n] \sum_{m\ge 1} (-1)^{m} \frac{1}{m} z^m \exp(mz) = [z^n] \log\frac{1}{1+z\exp(z)}.$$