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I need some help figuring out how to do this problem:

Prove that for every natural number n, [tex](n!)^{2} > n^n[/tex]

By rewriting this as

[tex](n!)^{\frac{1}{n}} > \sqrt{n}[/tex]

I can see that I have to prove that given a sequence of natural numbers {1, 2, 3, ...., n}, the geometric mean of n numbers of this sequence is greater than the geometric mean of its extreme terms, i.e. 1 and n. I get stuck however, when I attempt a noninductive proof (even an inductive proof for that matter involves handling binomial expressions which are large but it is perhaps still possible). I don't see how to go on now that I have the rearranged inequality even though in its present form and significance, it looks pretty obvious.

I would be grateful if someone could offer a suggestion to help me analyze this.

Thanks and cheers

Vivek

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# Square of n factorial is greater than n to the power n

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