TITLE: Show by induction that $2!4!6!...(2n)! \geq ((n+1)!)^n$
QUESTION [7 upvotes]: Show by induction that $2!4!6!...(2n)! \geq ((n+1)!)^n$
I stuck at $((n+1)!)^n (2(n+1))! \geq ((n+1+1)!)^{n+1}$, but cant progress to next step
It will be great in someone can demonstrate how to solve this problem. Thanks

REPLY [2 votes]: In order to prove$$ \prod_{i=1}^{n+1}(2i)!\ge((n+2)!)^{n+1} \tag{1}$$ you need to use directly your assumption $$ \prod_{i=1}^{n}(2i)!\ge((n+1)!)^n.$$
Indeed, note that $(1)$ is equivalent to $$ \prod_{i=1}^{n}(2i)!\cdot(2n+2)!\ge((n+1)!)^n\cdot(n+2)^n\cdot(n+2),$$ so it is implied by $$ (2n+2)!\ge(n+2)^{n+1} \\ (n+1)!\cdot\underbrace{(n+2)(n+3)(n+4)\ ... (2n)(2n+1)(2n+2)}_{n+1 \ \text{factors}}\ge\underbrace{(n+2)(n+2)\ ... (n+2)}_{n+1 \ \text{factors}}. $$ Hence if we let $\displaystyle k=\frac{\prod_{i=2}^{n+2}(n+i)}{(n+2)^{n+1}}>1$,  dividing both sides by $(n+2)^{n+1}$ we get $$k(n+1)! \ge 1,$$ which is obviously true for any $n$.