Prove that $2^n\le n!$ for all $n \in \mathbb{N},n\ge4$ The problem i have is:
Prove that $2^n\le n!$ for all $n \in \mathbb{N},n\ge4$
Ive been trying to use different examples of similar problems like at:
http://web.cacs.louisiana.edu/~mgr/261/induction.html
First i show the base case $n=4$ is true.
Then assuming $2^k\le k!$ for some $k \in \mathbb{N},n\ge4$
For $k+1$ we have $2^{k+1}\le (k+1)!$
Rewritten as $2\cdot2^k\le k!\cdot(k+1)$
Can you not simply say $2^k\le k!$ from the inductive hypothesis, and $2\lt4\le k\lt k+1$ proving the induction step?
I am having trouble following some of what seems to me like unnecessary steps like in the example, but feel like what i did above is wrong as im of course just learning how to use induction.

You can compare growth rates:
$$\frac{n!}{(n-1)!} = n,$$
$$\frac{2^n}{2^{n-1}} = 2 $$
This shows that for $n > 2$, as soon as $2^n \geq n!$, you'll have $2^N > N!$ for all $N > n$.
Sometimes it might be better to do $u_{n} - u_{n-1}$ instead of $\frac{u_n}{u_{n-1}}$.
You can also compare derivatives ($\frac{dy}{dx}$) or derivatives of logs ($\frac{d\log y}{dx}$).
