Problem 1006

[Proof] Prove by mathematical induction on $m$
$$S_{m, n}=(-1)^{m} \frac{(n+m)!}{n!} \text {. }$$
(1) When $m=1$,
$$\begin{aligned} S_{1, n} & =1-\frac{(n+2)!}{n!(n+1)} \\ & =1-(n+2) \\ & =-\frac{(n+1)!}{n!} . \end{aligned}$$

Thus, when $m=1$, (1) holds.
(2) Assume (1) holds for $m$, then for $m+1$,
$$\begin{aligned} S_{m+1, n} & =S_{m, n}+(-1)^{m+1} \frac{(n+m+2)!}{n!(n+m+1)} \\ & =(-1)^{m} \frac{(n+m)!}{n!}+(-1)^{m+1} \frac{(n+m)!(n+m+2)}{n!} \\ & =(-1)^{m+1} \frac{(n+m)!}{n!}[(n+m+2)-1] \\ & =(-1)^{m+1} \frac{(n+m+1)!}{n!}, \end{aligned}$$

Thus, (1) holds for $m+1$.
Therefore, (1) holds for all $m \in \mathbb{N}$.
Since $C_{n+m}^{m}$ is a natural number, then
$$\begin{aligned} S_{m, n} & =(-1)^{m} \frac{(n+m)!}{n!m!} \cdot m!. \\ & =(-1)^{m} C_{n+m}^{m} \cdot m! \end{aligned}$$

Thus, $S_{m, n}$ is divisible by $m!$.
When $n=2, m=3$,
$$S_{m, n}=S_{3,2}=-60$$

is divisible by $m!=3!=6$, but not by $m!(n+1)=3!(2+1)=18$.