Show by induction that $P_n : (2n-1)! \leq n^{2n-1}$.
My attempt at solution:
$P_n : (2n-1)! \leq n^{2n-1} $.
Check $P_1$: LHS = $1$ and RHS = $1$ $\implies$ $P_1$ is true.
Suppose $P_n$ is true; given that $(2n-1)! \leq n^{2n-1}$, then:
$(2(n+1) - 1)!$
$= (2n+1).(2n).(2n-1)!$
$\leq (2n+1).(2n).n^{2n-1}$
$= 2.(2n+1)n^{2n}$
$\leq 2.(2n+2)n^{2n}$
$= 4.(n+1).n^{2n}$
$\leq 4.(n+1).(n+1)^{2n}$
$ = 4.(n+1)^{2n+1}$
$ \leq 4.(n+1)^{2(n+1)-1}$
This is not quite $P_{n+1}$. Can anyone do better?
You can use the fact that $(1+\frac{1}{n})^n\ge 2$ for all $n\ge 1$.
You correctly obtained $4(n+1)n^{2n}$ and you need to show this is less than $(n+1)^{2n+1}$.
$$(n+1)^{2n+1}=(n+1)n^{2n}(1+\frac{1}{n})^{2n}$$
Over to you!