Show by induction that $2!4!6!...(2n)! \geq ((n+1)!)^n$ 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 
 A: Assume $\prod_{i=1}^n (2i)! \ge ((n+1)!)^n$ .
Then 
$$\prod_{i=1}^{n+1} (2i)! = \left((2n+2)!\right)\prod_{i=1}^{n} (2i)! \ge \left((2n+2)!\right)((n+1)!)^n = \frac{(2n+2)!}{(n+1)!} ((n+1)!)^{n+1}$$
Now, can you see why $\frac{(2n+2)!}{(n+1)!} \ge (n+2)^{n+1}$?
Hint: How many terms are in each? What is the relationship between each term of the lefthand product and any term of the righthand product?
A: 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 $n$.
