It isn’t actually true that $n!=(2\pi n)^{1/2}\left(\frac{n}e\right)^n\;:$ the righthand side is only approximately equal to $n!$. What is true, however, is that $\lim_{n\to\infty}\frac1{n!}\cdot(2\pi n)^{1/2}\left(\frac{n}e\right)^n=1\;,\tag{1}$ which would be good enough for your purposes if you were to approach the problem using Stirling’s approximation. That, however, is doing it very much the hard way. Here’s a much easier approach.
You want to show that $\lim_{n\to\infty}\frac{n!}{2^n}=\infty\;.$ The fraction has $n$ factors in both the numerator and the denominator, so you can write it as $\frac{n!}{2^n}=\prod_{k=1}^n\frac{k}2=\frac12\cdot\frac22\cdot\frac32\cdot\ldots\cdot\frac{n-1}2\cdot\frac{n}2\;.$ Call this product $f(n)$. By actual calculation $f(4)=3/2$. Now suppose that $n>4$; then
$f(n)=f(4)\prod_{k=5}^n\frac{k}2=\frac32\cdot\underbrace{\frac52\cdot\frac62\cdot\ldots\cdot\frac{n-1}2\cdot\frac{n}2}\;.$ Can you see why this is greater than $2^{n-4}$, and why that gives you the desired result?