Prove that $(n!)^2 ≥ n^n$ using mathematical induction

Question

1°

$n_0=1$

$(1!)^2 \ge 1^1$

$1\ge1$

2°

$k \ge n_0$

assumption: $$(k!)^2 \ge k^k$$

and for k+1: $$((k+1)!)^2 \ge (k+1)^{k+1}$$

I also noticed that: $$((k+1)!)^2 = (k!)^2 * (k+1)^2$$

Answer 1

Continuing from your steps, we are just required to prove

$$k^k \geq (k+1)^{k-1} \Leftrightarrow k^{k-1} \cdot k \geq (k+1)^{k-1}\Leftrightarrow (\frac{k}{k+1})^{k-1} \geq \frac{1}{k}$$

Using Bernouli's inequality, we have

$$(\frac{k}{k+1})^{k-1}=(1-\frac{1}{k+1})^{k-1} \geq 1-\frac{k-1}{k+1}=\frac{2}{k+1}$$

So, we are just left to prove $$\frac{2}{k+1} \geq \frac{1}{k}\Rightarrow 2k \geq (k+1) \Rightarrow k \geq 1$$ which is obvious.

Answer 2

A direct approach may be simpler: We have $$(n!)^2=\prod_{k=1}^nk\cdot\prod_{k=1}^{n}(n+1-k)=\prod_{k=1}^nk(n+1-k)\ge\prod_{k=1}^nn $$ because $k(n+1-k)=-k^2+(n+1)k$ is a "downward" quadratic, hence is minimal at the boundary places $k=1$ and $k=n$.