Considering the following summation of series: $$S_n=\sum_{k=0}^{n}(-1)^k{{n}\choose{k}}\sum_{m=0}^{k}(-1)^m\frac{k!}{(k-m)!}b^{-m},$$ where $n$ is a non-negative integer, and $b$ is a known non-zero constant.
I computed manually and got $$S_1=b^{-1}, S_2=2b^{-2}, S_3=6b^{-3}.$$ Then I set a hypothesis of $S_n$: $$S_n=n!\cdot b^{-n}.$$ However, I couldn't prove whether it is correct or not. Could someone help me? Thanks very much indeed!
Change the order of summation, then shift the summation index $k$ to $k-m$ and then use the binomial theorem:
\begin{align*}&\hskip-2cm\sum_{k=0}^{n}(-1)^k{{n}\choose{k}}\sum_{m=0}^{k}(-1)^m\frac{k!}{(k-m)!}b^{-m}\\&= \sum_{0\le m\le k \le n}(-1)^k \frac{n!}{k!(n-k)!}(-1)^m\frac{k!(n-m)!}{(k-m)!(n-m)!}b^{-m}\\ &= \sum_{m=0}^{n}(-1)^mb^{-m}\frac{n!}{(n-m)!}\sum_{k=m}^{n}(-1)^k\binom{n-m}{k-m}\\ &=\sum_{m=0}^{n}b^{-m}\frac{n!}{(n-m)!}\sum_{k=0}^{n-m}(-1)^{k}\binom{n-m}{k}\\ &=\sum_{m=0}^{n}b^{-m}\frac{n!}{(n-m)!}(1-1)^{n-m}\\ &=\sum_{m=0}^{n}b^{-m}\frac{n!}{(n-m)!}\delta_{n,m}\\ &=n!b^{-n}.\end{align*}