As the title states, I am trying to prove that $\sum_{k=0}^n \frac{(-1)^k}{n!}k^n\binom{n}{k} = (-1)^n$. I verified (numerically) that this is true, and you can find the sequence $\sum_{k=0}^n (-1)^k k^n \binom{n}{k} = (-1)^n n!$ on https://oeis.org/A133942. I found that induction was not a good way to approach this problem due to the existence of the $k^n$ and $\binom{n}{k}$ inside of the summand. Since there is an $n!$ in the formula, I tried to apply the Cauchy Integral formula $$f^{(n)}(z_0) = \frac{n!}{2\pi i}\int_\gamma \frac{f(z)}{(z - z_0)^{n + 1}}\:dz,$$ but I appear to be having trouble getting the whole thing to work out. I have also tried expanding the sum using the identity $k\binom{n}{k} = n\binom{n-1}{k-1}$, but that did not yield anything useful either. Any help would be appreciated. Thank you!
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1You may tackle the problem by just considering how $\delta^n$ acts on $p(x)=x^n$, with $\delta$ being the forward difference operator, sending $p(x)$ into $p(x+1)-p(x)$. – Jack D'Aurizio May 30 '22 at 19:27
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1This can be proved in many ways. For example, you may investigate the power series expansion of both sides of $$(e^x - 1)^n=\sum_{k=0}^{n} (-1)^{n-k}\binom{n}{k}e^{kx},$$ or use the inclusion-exclusion principle to count the number of bijections from ${1,2,\ldots,n}$ to ${1,2,\ldots,n}$. – Sangchul Lee May 30 '22 at 19:29
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Ah, thank you all! That was all that I needed. JackD'Aurizio, thank you for that insight: I arrived at this formula from some computations using forward differences, so that expansion makes a lot of sense. Sangchul, thank you for the quick formulation: I tried something similar but obviously missed the expansion. And TheSilverDoe, thank you for the reference: I tried to find that on the site, but I obviously missed it. I will close the question now. – peabody May 30 '22 at 19:37
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No, don't close the question. This is a worthwhile question that provides a good reference to others. Simply leave the posting, as is. – user2661923 May 30 '22 at 21:22