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.

An easy and intuitive solution. 
Write $k>3$ in place of $k\geq4$. 
One can easily prove the base case. 
Now Assume that $2^k\leq k!$ 
So lets prove that $2^{k+1}\leq (k+1)!$
$2^k.2\leq (k+1)k!$ 
Multiply both sides by $-1$ and flip the sign.
$-2^k.2> -(k+1)k!$ 
$k!\geq 2^k$ ->Assumption 
So $-2^k.2> -(k+1)2^k$ 
$-2>-k-1$ Divide both sides by $2^k$ and flip the sign 
So k>1 
Hence proved.
