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Can somebody in elementary way show that $(n!)^{2\over n+1}>n-1$ for only finitely many $n\in\mathbb N$? I need to prove this to be able to prove something else.

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I don't think it's true. It's known that $\log(n!)$ is very roughly $n\log n$, so the log of the left side is very roughly $2\log n$, while the log of the right side is roughly $\log n$, which is smaller. This suggests the inequality holds for all but finitely many $n$.

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But if $\log(n!)$ is very roughly $n\log n$ than $\log(n!)^{2\over n+1}$ is very roughly ${2n\over n+1}\log n$ which is roughly, as you said, $2\log n$ and it sure is greater than $\log (n-1)$. But I think that for large $n$ the $\log (n!)$ behaves more like ${n\over2}\log n$ so the inequality could be wrong for large $n$. Can we be more precise? –  A.P. Feb 8 '13 at 11:40
    
If you are familiar with Stirling's approximation, you'll be able to work out that it's $n\log n$, not $(n/2)\log n$. More accurate is $(n+(1/2))\log n-n$, but for our purposes that's still $n\log n$. Try calculating $\log(100!)$, and see whether it's closer to $50\log100$ or $100\log100$ (or $(100.5)\log100-100$). All logs here are natural, naturally. –  Gerry Myerson Feb 8 '13 at 12:28
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