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Is it true that $\sqrt{n!}\notin\mathbb{Z}$ for $n>1$? This is what I did: By induction: for $n=2$ its trivial ($\sqrt{2}$ is irrational). Suppose its true for some $n\in\mathbb{N}$, then the primefactorization has to be $n=p_1^{a_1}\cdots p_m^{a_m}$ for some $m\in \mathbb{N}$ where there has to be a $a_i$ such that $2\nmid a_i $ and $1\leq i\leq m$. If $n+1$ is prime, then we are done. suppose its not prime, then $n+1=q_1^{b_1}\cdots q_l^{b_l}$ for some $l\in \mathbb{N}$. From here on I got stuck. First: is the statement actually true (give counterexample)? If its true, can you give me hints? Thanks.

Andrés E. Caicedo
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    Hint: Bertrand's Postulate/Chebyshev's Theorem. – Daniel Fischer Aug 26 '13 at 15:20
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    See also http://math.stackexchange.com/questions/31973/n-is-never-a-perfect-square-if-n-geq2-is-there-a-proof-of-this-that-doesn and http://math.stackexchange.com/questions/52844/elementary-proof-regarding-the-largest-prime-in-the-factorization-of-n?lq=1 – Jonas Meyer Aug 26 '13 at 15:21
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    Yeah, Betrand is the first thing that came to my mind. Not sure if there is a more direct proof - Bertrand is a big sledgehammer to wield. – Thomas Andrews Aug 26 '13 at 15:22
  • Can we do this by induction? I have a feeling we can and it's going to be rigorous. I'm off to try! – Sum-Meister Feb 06 '17 at 14:13

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