Proving that $\prod_{i=1}^n (2i)! \ge ((n+1)!)^{n}$ using strong induction We are not only confused about how many initial cases to prove, but also how to use the inductive step. For proving the case $n=k+1$,  I have tried to factor $((k+2)!)^{k+1}$ into terms involving $k-1$ because I know that I need to use $n\le k$. However, I am unsure of where to go from there or even if that is the correct route to take. I would also appreciate any general advice on using strong induction in proofs. Thank you. 
 A: You need just one initial case, $n=1$. However, you don’t need strong induction. For the induction step assume that
$$\prod_{i=1}^k(2i)!\ge((k+1)!)^k\;,\tag{1}$$
and you want to show that
$$\prod_{i=1}^{k+1}(2i)!\ge((k+2)!)^{k+1}\;.\tag{2}$$
The natural first step is to split the lefthand side of $(2)$ into the part that you know about and the new factor:
$$\prod_{i=1}^{k+1}(2i)!=\left(\prod_{i=1}^k(2i)!\right)(2k+2)!\;.\tag{3}$$
In other words, in going from $(1)$ to $(2)$ we’ve multiplied the lefthand side by $(2k+2)!$; if we’ve multiplied the righthand side of $(1)$ by the same amount or less, $(2)$ will certainly be true. So what have we multiplied $((k+1)!)^k$ by to get $((k+2)!)^{k+1}$?
$$\frac{((k+2)!)^{k+1}}{((k+1)!)^k}=\left(\frac{(k+2)!}{(k+1)!}\right)^k(k+2)!=(k+2)^k(k+2)!\;,$$
so to get $(2)$ from $(1)$ we’ve multiplied the lefthand side by $(2k+2)!$ and the righthand side by $(k+2)^k(k+2)!$. If we can show that
$$(2k+2)!\ge(k+2)^k(k+2)!\;,\tag{4}$$
we’ll be in business: $(1)$ will then imply $(2)$, and we’ll have carried out the induction step.
HINT: $(4)$ is equivalent to the inequality
$$\frac{(2k+2)!}{(k+2)!}\ge(k+2)^k\;;$$
do all of the cancelling that you can on the left and compare the product that’s left with the righthand side.
