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Just what is asked in the title, with $k$ an integer such that $0\le k<n$.

Alessandro Della Corte
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    I don't know that it has a name, but we have a proof of it: https://math.stackexchange.com/questions/1617965 – 311411 Sep 26 '21 at 15:16
  • Thanks, but I do have a proof. I just wanted to know in which context does this usually pop up and possibly a name associated with it. – Alessandro Della Corte Sep 26 '21 at 15:17
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    There are an unbounded number of identities. Only a few of them are named. This is not one of them, but perhaps it deserves to be named. – Somos Sep 26 '21 at 16:53
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    I hereby name the identity Graviton's Identity. Those not in favour I will be dueling at sundown. – Graviton Sep 26 '21 at 18:33

2 Answers2

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The sum $\sum _{i=0}^n (-1)^i i^k \binom{n}{i}$ is known as $(-1)^n n! \mathcal{S}_k^{(n)}$, involving the so-called Stirling number of second kind, denoted as $\mathcal{S}_k^{(n)}$.

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Consider the (forward) finite Difference of a function, being defined as $$ \Delta f(x) = f(x + 1) - f(x) $$ and its iteration $$ \Delta ^n f(x) = \Delta \left( {\Delta ^{n - 1} f(x)} \right) = = \sum\limits_{\left( {0 \le } \right)k\left( { \le n} \right)} {\left( { - 1} \right)^{n - k} \left( \begin{array}{c} n \\ k \\ \end{array} \right)f(x + k)} $$

Then $$ \left. {\Delta ^n x^m } \right|_{x = 0} = \sum\limits_{\left( {0 \le } \right)k\left( { \le n} \right)} {\left( { - 1} \right)^{n - k} \left( \begin{array}{c} n \\ k \\ \end{array} \right)k^m } \left. {\Delta ^n x^m } \right|_{x = 0} = \left( { - 1} \right)^n \sum\limits_{\left( {0 \le } \right)k\left( { \le n} \right)} {\left( { - 1} \right)^k \left( \begin{array}{c} n \\ k \\ \end{array} \right)k^m } $$ So what you are having is $(-1)^n$ times the $n$-th difference of $x^k$ computed at $x=0$, hence the relation with the Stirling N. .

G Cab
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