A closed form for $\sum_{k=0}^{n}(-1)^k\frac{k^p}{k!(n-k)!}$

Question

Is there a closed form for $$ \sum_{k=1}^n (-1)^k\frac{k^p}{k!(n-k)!},\quad n=0,1,2\ldots,\,p=0,1,2\ldots. $$ I tried to identify the sum with Stirling numbers...

Answer 1

From my comment and Pedro Tamaroff's suggestion, let $k \leftarrow n - k$.

Then we get $$\dfrac{1}{n!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{n-k} (n-k)^p = \dfrac{1}{n!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} (n-k)^p = (-1)^n{ p \brace n} $$

This result only holds if $p > 0$. If $p = 0$, $$\dfrac{1}{n!} \sum_{k=1}^n (-1)^{k} \binom{n}{k} = -\dfrac{1}{n!} $$