Mathematical induction $n! > n^{\frac n 2}$

Question

This problem from Lyashko I.I. et al. "AntiDemidovich: higher mathematic reference book", vol. $1$, ch. $1$ "Introduction to analysis" (ISBN $978-5-9710-7384-0$)

Problem 31.a) Prove by mathematical induction:

$$ n! > n^{n/2} $$

Gave up on proof for $n! > n^{\frac n 2}$ for $n+1$:

$$(n+1)! = n! \cdot(n+1) > (n+1) \cdot n^{\frac n 2} = n^{\frac{n+2}{2}} + n^{\frac n 2} = {???}$$

How can I solve this problem?

Answer 1

Assume $n!^2>n^n$ and lets consider following steps $$(n+1)!^2=n!^2(n+1)^2>n^n(n+1)^2$$ now we want $n^n(n+1)^2>(n+1)^{n+1} \Leftrightarrow \frac{n^n}{(n+1)^n}>\frac{1}{n+1}$. Can you complete from here?

Answer 2

The easiest way is not to use induction at all. $$n!^2=\prod_{i=0}^{n-1}(i+1)\times\prod_{i=0}^{n-1}(n-i)=\prod_{i=0}^{n-1}(i+1)(n-i)\\=\prod_{i=0}^{n-1}(n+(n-i-1)i)>\prod_{i=0}^{n-1}n=n^n.$$