Prove using induction $ 2!4!...(2n)!>[(n+1)!]^n$

Question

Prove using induction $ 2!4!...(2n)!>[(n+1)!]^n$

I have tried out a lot of ways to solve this, one of them being taking all the 2's out:

$ 2!4!6!...(2n)!$
$ =2^{1+2+3+...+n}(1!2!3!...n!)$
$ =2^{\frac{n(n+1)}{2}}(1!2!3!...n!)$

But I have not been able to solve this problem. It would be great if someone could help me with this.

Answer 1

Directly, we have $$\begin{align*} (2n+2)! &= (n+1)! \prod_{k=1}^{n+1} (n+k+1) \\ &> (n+1)! \prod_{k=1}^{n+1} (n+2) \\ &= (n+1)! (n+2)^{n+1} \\ &= \frac{(n+2)!^{n+1}}{(n+1)!^n} \end{align*}$$ which establishes the induction step: we then have $$\prod_{m=1}^{n+1} (2m)! = (2n+2)! \prod_{m=1}^n (2m)! > \frac{(n+2)!^{n+1}}{(n+1)!^n} \cdot (n+1)!^n = (n+2)!^{n+1},$$ so if the claim is true for $n$, it is also true for $n+1$.

Answer 2

Let $$a_n = \frac{\prod_{k=1}^{n} (2k)!}{((n+1)!)^n}$$

Note that $a_2 = \frac{4}{3}>1$. We will now prove that $a$ is increasing (this is kinda-sorta induction, and would rely on induction if we were to be completely rigorous). This increasing-ness is equivalent to

$$a_{n-1}<a_n$$

$$\frac{a_n}{a_{n-1}}>1$$

$$\frac{(2n)!}{\frac{((n+1)!)^n}{n!^{n-1}}}>1$$

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

To prove this, note that this taken at $n=2$ is greater than $1$. We will now prove that this is increasing. This is equivalent to

$$\frac{(2n)!}{n!(n+1)^n}\frac{(n-1)!n^{n-1}}{(2n-2)!}>1$$

$$\frac{2n(2n-1)n^n}{n^2(n+1)^n}>1$$

$$\left(4-\frac{2}{n}\right)\left(\frac{n}{n+1}\right)^n>1$$

$$4-\frac{2}{n}>\left(\frac{n+1}{n}\right)^n$$

Note that the LHS in this inequality is always $>3$, and the RHS is always less than $e<3$. Thus, this inequality holds, finishing the proof.

Note that, at $n=1$, the two sides of the original inequality are equal.