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I want to prove that for all $0 \leq l \leq n$, $$ \displaystyle\sum_{k=0}^n \displaystyle\binom{n}{k} \dfrac{(-1)^{n-k}k^l}{l!} = \begin{cases} 0 \text{ if } l<n \\ 1 \text{ if } l=n \end{cases}$$

I know that $\displaystyle\sum_{k=0}^n \displaystyle\binom{n}{k}(-1)^{n-k} = 0$ I tried to use summation by parts(Abel lemma) but I wasn't able to conclude. Thanks for your help!

Lasky
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  • Proof by inference? – A rural reader Mar 27 '24 at 16:58
  • I'm confused, either it's $l^k$ in the question or $(n-k)^l$ in the comment, right? – Matija Mar 27 '24 at 17:08
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    Edit of previous comment: This can be rewritten as $$\sum(−1)^k\binom nk (n−k)^l=\begin{cases}0&0\leq l<n\n!&l=n\end{cases}$$ and the left side can be seen as the number of functions ${1,2,…,l}\to{1,…,n}$ that are onto, using inclusion-exclusion – Thomas Andrews Mar 27 '24 at 17:15
  • Specifically, if $A$ is the set of all functions $f:{1,\dots,l}\to{1,\dots,n}$ and $A_i$ is the set of all such functions which don't have $i$ in the image, then the set of onto maps is $A\setminus(A_1\cup \cdots\cup A_n),$ and count that set using inclusion-exclusion. – Thomas Andrews Mar 27 '24 at 17:20

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