How to prove $(n+1)! = O(2^{(2^n)})$
Well, since this upper bound is not nearly tight, you can just use basic transformations to get

$$(n+1)!&lt; (n+1)^{n}=2^{\log (n+1)^{n}}= 2^{n\log (n+1)}=O(2^{n\log n})\subset O(2^{(2^n)}). $$