Show $n!=\omega(2^n)$ using Stirling's Approximation

Question

I'm faced with a question to show that $n!=\omega(2^n)$ using Stirling's Approximation $n!\sim \sqrt{2\pi n}\cdot e^{-n}n^{n}$.

Showing $n!=\omega(2^n)$ is equivalent to showing that $\lim_{n\to\infty} {\frac{n!}{2^n}}=\infty$.

I tried: $$\begin{align*} \lim_{n\to\infty}{\frac{n!}{\sqrt{2\pi n}\cdot e^{-n}n^{n}}}&=1\\ {\frac{\lim_{n\to\infty}{n!}}{\lim_{n\to\infty}{\sqrt{2\pi n}\cdot e^{-n}n^{n}}}}&=1\\ \lim_{n\to\infty}{n!}&=\lim_{n\to\infty}{\sqrt{2\pi n}\cdot e^{-n}n^{n}}\\ \lim_{n\to\infty} {\frac{n!}{2^n}}&=?\\ {\frac{\lim_{n\to\infty}{\bigg[\sqrt{2\pi n}\cdot e^{-n}n^{n}\bigg]}}{\lim_{n\to\infty}{2^n}}}&=?\\ {\frac{\lim_{n\to\infty}{\bigg[\sqrt{2\pi n}\bigg]\cdot \lim_{n\to\infty}{\bigg[e^{-n}\bigg]}\cdot \lim_{n\to\infty}{\bigg[n^{n}\bigg]}}}{\lim_{n\to\infty}{2^n}}}&=?\\ \end{align*}$$

At this point I feel I'm wrong and can't continue since $\lim_{n\to\infty}{\bigg[e^{-n}\bigg]}=0$, so I can't get the result I want. Any ideas?

Answer 1

Hint:

If $a_k$ is a non-decreasing positive sequence, then

$$\lim_{n\to\infty}(1+a_n)^n=\infty$$

Answer 2

Stiling's Formula is

$$n!=\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac1n\right)\right)$$

Therefore, we can write

$$\frac{n!}{2^n}=\frac{\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\left(1+O\left(\frac1n\right)\right)}{2^n}=\sqrt{2\pi n}\left(\frac{n}{2e}\right)^n\left(1+O\left(\frac1n\right)\right)$$

Now, for $n\ge 6$, $\frac{n}{2e}>1$. Therefore, $\left(\frac{n}{2e}\right)^n\to \infty$ as $n\to \infty$, from which it is easy to see that

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