limits - Finding $lim_{n to infty} frac{sqrt{n!}}{2^n}$

I'm looking for a way to find this limit:

$\lim_{n \to \infty} \frac{\sqrt{n!}}{2^n}$

I think I have found that it diverges, by plugging numbers into the formula and "sandwich" the result. However I can't find way to prove it.

I know that $n! \approx \sqrt{2 \pi n}(\frac{n}{e})^n$ when $n \to \infty$. (Stirling rule I think)

However, I don't know how I could possibly use it. I mean, I tried using the rule of De l'Hôpital after using that rule, but I didn't go any further.

Answer

$$\frac{\sqrt{n!}}{2^n}\sim\frac{(2\pi\,n)^{1/4}\Bigl(\dfrac{n}{e}\Bigr)^{n/2}}{2^n}=(2\pi\,n)^{1/4}\Bigl(\frac{\sqrt n}{2\sqrt e}\Bigr)^{n}.$$ Since $\dfrac{\sqrt n}{2\sqrt e}$ and $n^{1/4}$ converge to $\infty$, so does $\dfrac{\sqrt{n!}}{2^n}$.