Prove by induction that $ 2!4!..(2n)! > ((n+1)!)^n $

Question

Prove by induction that $ 2!4!..(2n)! > ((n+1)!)^n $.
Attempt :
It is clearly true for $n=2$.
Let it be true for $n=m$.
Therefore, $ 2!4!..(2m)! > ((m+1)!)^m $
=> $ 2!4!..(2m)!(2(m+1))! > ((m+1)!)^m . {2(m+1)}! $

It will be done if I can prove that :

$ ((m+1)!)^m . (2(m+1))! > ((m+2)!)^(m+1) = ((m+1)!)^m . (m+2)^m.(m+2)!$

which boils down to proving :

$ (2(m+1))! > (m+2)^m.(m+2)! $
I am unable to prove it. If i try to prove it by induction, it leads to more expressions that are needed to be proved.

Answer 1

\begin{align} (2(m+1))! &= (2m+2)(2m+1) \cdots (m+3)\cdot(m+2)!\\ &> (m+2)(m+2) \cdots (m+2) \cdot (m+2)!\\ &= (m+2)^m (m+2)! \end{align}

Answer 2

\begin{align*} (2(m+1))!&=(m+2)!\underbrace{(m+3)\cdots(2(m+1))}_{m\text{ terms}}\\ &>(m+2)!(m+2)^{m} \end{align*}

Answer 3

Base case: $n=2$. Since $2!4! > ((2+1)!)^2 \implies 48 > 36$, the base case is true.

Inductive step: Suppose that $2!4!..(2n)! > ((n+1)!)^n$. We need to show that $2!4!..(2n)!(2(n+1))! > ((n+2)!)^{n+1}$

Begin with

$2!4!..(2n)! > ((n+1)!)^n$

Then multiply both sides by $(2(n+1))!$

$2!4!..(2n)!(2(n+1))! > ((n+1)!)^n(2(n+1))!$

$2!4!..(2n)!(2(n+1))! > ((n+1)!)^n\underbrace{(2n+2)(2n+1)(2n)...(n+3)}_{n\text{ terms}}(n+2)!$

Decrease all the terms to $n+2$ in the bracket, as the inequality will still hold:

$2!4!..(2n)!(2(n+1))! > ((n+1)!)^n\underbrace{(n+2)(n+2)(n+2)...(n+2)}_{n\text{ terms}}(n+2)!$

From here, simplify:

$2!4!..(2n)!(2(n+1))! > ((n+1)!)^n(n+2)^n(n+2)!$

$2!4!..(2n)!(2(n+1))! > ((n+2)!)^n(n+2)!$

$2!4!..(2n)!(2(n+1))! > ((n+2)!)^{n+1}$