Prove $(n!) ^2 > n^n$ for all $n > 3$ using Induction

Question

Prove $(n!) ^2 > n^n$ for all $n > 3$ by induction.
I know that we will have to use Binomial theorem somehow but I can't figure out how?
Please provide some hint.

Answer 1

HINT

Proceed as usual by

• base case
• induction step: $$(n!)^2>n^n \implies ((n+1)!)^2>(n+1)^{n+1}$$

then consider

$$((n+1)!)^2=(n+1)^2(n!)^2\stackrel{Ind. Hyp.}>(n+1)^2n^n\stackrel{?}>(n+1)^{n+1}$$

therefore to conclude we need to prove the last inequality that is

$$(n+1)^2n^n\stackrel{?}>(n+1)^{n+1}$$

Answer 2

Hint for the induction step:

$(n+1)^2n^n>(n+1)^{n+1} \iff (1+1/n)^n<n+1$.

Answer 3

On the left side, $$\frac{(n+1)!^2}{n!^2}=(n+1)^2$$ and on the right side

$$\frac{(n+1)^{n+1}}{n^n}=(n+1)\left(1+\frac1n\right)^n=(n+1)\left(1+\frac nn+\frac{n(n-1)}{n^22!}+\frac{n(n-1)(n-2)}{n^33!}+\cdots\right).$$

Then

$$\left(1+\frac1n\right)^n=1+\frac nn+\frac{n(n-1)}{n^22!}+\frac{n(n-1)(n-2)}{n^33!}+\cdots<1+1+\frac12+\frac1{3!}+\cdots<e.$$