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Perform a Taylor expansion of $e^x$ at $x=0$, then $$e^n=1+n+\frac{n^2}{2!}+\dots+\frac{n^n}{n!}+\int_0^{n}\frac{(n-t)^n}{n!}e^ndt,$$ we get $$(1+n+\frac{n^2}{2!}+\dots +\frac{n^n}{n!})e^{-n}=1-e^{-n}\int_0^{n}\frac{(n-t)^n}{n!}e^tdt=1-\frac{1}{n!}\int_0^nt^ne^{-t}dt.$$

Therefore, it suffices to prove that $\frac{1}{n!}\int_0^nt^ne^{-t}dt=\frac{1}{2}.$

How to get the desired outcome? I note $\int_0^{\infty}t^ne^{-t}dt=n!$... Can you give me some hints? Thanks in advance!

Ychen
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    This surely already exists here. My favorite proof of this uses the Central Limit Theorem. HERE: https://math.stackexchange.com/q/1599632/442 – GEdgar Sep 23 '23 at 05:32
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    It is not true that $\frac 1 {n!}\int_0^nt^ne^{-t}dt=\frac1 2$ but LHS tends to $\frac 1 2$. – geetha290krm Sep 23 '23 at 05:34
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    This question with many solutions, including $\frac{1}{n!}\int_n^\infty e^{-t},t^n,\mathrm{d}t$ ... https://math.stackexchange.com/q/160248/442 – GEdgar Sep 23 '23 at 05:38
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    It's worth reading the answers to this on Quora. Especially Alon Amit's. https://www.quora.com/How-do-you-prove-that-displaystyle-lim_-limits-n-to-infty-e-n-sum_-k-0-n-frac-n-k-k-frac-1-2 – Blitzer Sep 23 '23 at 05:42

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