# How can I prove $(n^2)!$ greater growth than $(n!)^n$ [duplicate]

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

## 3 Answers

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 $$(n!)^n<(n^2)!;\;\forall n\in\mathbb{N},n>1$$

• I downvoted (sorry) - I think this answer is making matters much more complicated than they need to be. – MCT Nov 26 '20 at 16:00

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.

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?

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