Proof by contradiction that $n!$ is not $O(2^n)$ I am having issues with this proof: Prove by contradiction that $n! \ne O(2^n)$. From what I understand, we are supposed to use a previous proof (which successfully proved that $2^n = O(n!)$) to find the contradiction.
Here is my working so far:
Assume $n! = O(2^n)$. There must exist $c$, $n_{0}$ such that $n! \le c \cdot 2^n$. From the previous proof, we know that $n! \le 2^n$ for $n \ge 4$.
We pick a value, $m$, which is gauranteed to be $\ge n_{0}$ and $\ne 0$. I have chosen $m = n_{0} + 10 + c$.
Since $m > n_0$:
$$m! \le c \cdot 2^m\qquad (m > n \ge n_0)$$
$$\dfrac{m!}{c} \le 2^m$$
$$\dfrac{1}{c} m! \le 2^m$$
$$\dfrac{1}{m} m! \le 2^m\qquad (\text{as }m > c)$$
$$(m - 1)! \le 2^m$$
That's where I get up to.. not sure which direction to head in to draw the contradiction.

In 
Factorial Inequality problem $\left(\frac n2\right)^n > n! > \left(\frac n3\right)^n$,
they have obtained
$$
n!\geq \left(\frac{n}{e}\right)^n.
$$
Hence 
$$
\lim_{n\rightarrow\infty}\frac{n!}{2^n}\geq\lim_{n\rightarrow\infty}\left(\frac{n}{2e}\right)^n=\infty.
$$
Suppose that $n!=O(2^n)$. Then there exist $C>0$ and $N_0\in \mathbb{N}$ such that 
$$\frac{
n!}{2^n}\leq C
$$
for all $n\geq N_0$. Lettting $n\rightarrow\infty$ in the aobve inequality we obtain
$$
\lim_{n\rightarrow\infty}\frac{n!}{2^n}\leq C,
$$
which is an absurd.
