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The following is what I've done, but I feel it is too simple. There must be some lapse in logic I'm not seeing.

Proof:

By the AM-GM inequality, $$\sqrt[n]{n!}≤\frac{1+2+...+n}n=\frac{\frac{n(n+1)}2}n=\frac{n+1}2$$ Since $\sqrt[n]{n!}≤\frac{n+1}2$, this implies that $n!≤(\frac{n+1}2)^n$.

I've looked at this thread, Proving that $n!≤((n+1)/2)^n$ by induction, and the proof with $i\cdot(n+1-i)≤(\frac{n+1}2)^2$ did not make sense to me. I did not see how it used the AM-GM inequality, or how it showed $n!≤(\frac{n+1}2)^n$

Joe Hunter
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    Your proof is fine, compare https://math.stackexchange.com/questions/992056/how-can-i-show-that-n-leqslant-fracn12n. – Martin R Oct 06 '17 at 17:16
  • The argument you mention is the AM-GM inequality tor two elements(GM² $\le$ AM²). – Bernard Oct 06 '17 at 17:45

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