Binomial identity

$$\displaystyle \sum_{k=1}^n(-1)^{n-k}\binom{n}{k}k^j=\begin{cases} n!, & j=n\\
0, & j=1,2,\ldots,n\end{cases}$$
Proof. Consider the number of surjective functions $f:\{1,2,\ldots,j\}\to\{1,2,\ldots,n\}$. If we count them using the inclusion-exclusion principle we get the sum in the problem. On the other hand this number is trivially $n!$ if $j=n$ and $0$ if $j<n$.
Remark 1. For $j>n$ we again find the number of surjections, which has a closed form form some $j$.
Remark 2. Interesting identity which represents the above identities in a unified manner is the following $$\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}(x+k)^n=n!$$ for all $x$ (in $\mathbb{R}$ for example, but any commutative ring suffices).

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