$\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!}(k+m)!=0$. I am working on showing an integral is zero, and I have simplified to the following
for $m&lt;n$.
 $\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!}(k+m)!$. 

From the few test cases I have done, the sum above always seems to be zero. How can I show it is zero for $m&lt;n$? What will it be when $m=n$?
Hint:

$$\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k(k+m)!}{k!}=m!\sum_{k=0}^n\binom{n}{k}\binom{k+m}{m}(-1)^k$$
$$=m![x^m]\sum_{k=0}^n\binom{n}{k}(1+x)^{k+m}(-1)^k$$

Where $[.]$ denotes the Coefficient operator. Can you proceed hereon?
\begin{align}
\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!}(k+m)! &amp; = \sum_{k=0}^n \binom{n}{k}(-1)^km!\binom{k+m}{k} \\
&amp;= m!\sum_{k=0}^n \binom{n}{k}\binom{-m-1}{k} \\
&amp;= m!\binom{n-m-1}{n} \\
&amp;= 0 \text{ when } n, m \in \textbf{N}.\
\end{align}
(If these steps seem a bit arcane, just know that they're all explained in Knuth's Art of Computer Programming, Vol. 1 where he gives a comprehensive list of ways to manipulate binomial coefficients.)