limit of sequence with factorial

How do you show that: $\lim\limits_{n\to \infty} \frac{\left(\frac{n}{2}\right)^{\frac{n}{2}}}{n!}=0$
using the squeeze theorem (I'd like to avoid using Stirling's formula, too). I tried rearranging it a bit into $\lim\limits_{n\to \infty} \frac{\left(\sqrt{n}\right)^{n}}{\left(\sqrt{2}\right)^{n}n!}$ , but i can't really figure out what to do next. Thanks!

Answer

The desired limit is equivalent to

$$lim_{n->\infty} {\frac{n^n}{(2n)!}}$$

Since

$$n^n < 2n*(2n-1)*...*(n+1)$$

we have the majorant

$$\frac{1}{n!}$$

which clearly tends to 0, if n tends to infinity.