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I tried take limit n goes to infinity. Wolfram solved the limit and $(n^2)!$ has greater growth rate but there is not step by step solution.

I think I can simplify $(n!)^n$ to $(n^n)^n$ But how can I prove $(n^2)! > (n^n)^n$

Wolfram link

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Use Stirling's approximation $$(n^2)!\sim \sqrt{2 \pi } e^{-n^2} n^{2 n^2+1}$$ and $$(n!)^n\sim (2 \pi n)^{n/2} e^{-n^2} n^{n^2}>e^{-n^2} n^{n^2+n}$$ Therefore $$(n!)^n<(n^2)!;\;\forall n\in\mathbb{N},n>1$$

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  • $\begingroup$ I downvoted (sorry) - I think this answer is making matters much more complicated than they need to be. $\endgroup$ – MCT Nov 26 '20 at 16:00
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You can write the terms as $$(n!)^n = \underbrace{1 \cdot \dotso \cdot 1}_{n\text{ times}} \cdot \underbrace{2 \cdot \dotso \cdot 2}_{n\text{ times}} \cdot \dotso \cdot \underbrace{n \cdot \dotso \cdot n}_{n\text{ times}}$$ and $$ (n^2)! = \underbrace{(1 \cdot \dotso \cdot n)}_{n \text{ terms}} \cdot \underbrace{((n+1) \cdot \dotso \cdot 2n)}_{n \text{ terms}} \cdot \dotso \cdot \underbrace{((n(n-1)+1) \cdot \dotso \cdot n^2)}_{n \text{ terms}}$$ You can look at the numbers as groups of $n$ numbers. You can see that the second term is always larger.

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Hint: Both $(n^2)!$ and $(n!)^n$ can be written as a product of a bunch of smaller terms. In fact, they can both be written as a product of $n^2$ terms. How do these terms compare to each other?

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  • $\begingroup$ How can I convert to smaller terms.I don't know @MCT $\endgroup$ – snowl Nov 25 '20 at 13:33

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