calculus - Prove that $lim_{n to infty} frac{n!}{2^{2^n}} = 0$.

Prove that $$\lim_{n \to \infty} \frac{n!}{2^{2^n}} = 0$$
I don't think I can use L'Hopital rule. I can't think of other ways. Do I need non-elementary results to prove this? I know Stirling's formula but I can't see how it helps. Hints appreciated.

Answer

Let $a_n=\frac{n!}{2^{2^n}}$. Then by using absurdly weak estimates

$$\frac{a_{n+1}}{a_n}=\frac{n+1}{2^{2^n}}<\frac{2^n}{2^{2^n}}< \frac{2^{2^n-1}}{2^{2^n}}=\frac12$$