How to prove $(n+1)! = O(2^{(2^n)})$

I am trying to prove $(n+1)! = O(2^{(2^n)})$. I am trying to use L'Hospital rule but I am stuck with infinite derivatives.

Can anyone tell me how i can prove this?

Well, since this upper bound is not nearly tight, you can just use basic transformations to get

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

StackExchange Computer Science Q#6075, answer score: 12