Proof by induction: $2^{n} < n!$ Prove that $2^{n} &lt; n!$ $\forall$ n > 4


$n=5:$ $$2^{5}&lt;5!$$

$$32 &lt; 120$$

This is true.

Now, after knowing it worked for $n$ we need to show it works for every other, so $n+1$:

$$2^{n+1} &lt; (n+1)!$$

$$2^{n} &lt; \frac{(n+1)!}{2}$$ 

We know from beginning that $2^{n} &lt; n!$

So we replace it with it here and then show:

$$n! &lt; \frac{(n+1)!}{2}$$

$$n! &lt; \frac{n!\cdot(n+1)}{2}$$

$$1&lt;\frac{n+1}{2}$$

Task say for all $n &gt; 4$ so the thing on the right side will really be greater than $1$.



I hope everything is ok?

Edit: The possible-duplicate-link didn't help me because I'm not really looking for a solution to the task. I'm rather interested in knowing if MY proof is correct.
I would simply note that for sufficiently large $n$, $2$ &lt; $n$ + $1$.  Then by induction, when $2^n$ &lt; $n$!, $2^n$ ${\times}$ $2$ &lt; $n$! ${\times}$ $2$ and $n$! ${\times}$ $2$ &lt; $n$! ${\times}$ $n + 1$, so by transitivity $2^n$ ${\times}$ $2$ &lt; $n$! ${\times}$ $n + 1$.
Are you stating that if $a&lt;b$ and $a&lt;c$ then $b&lt;c?$ But you have the correct answer. I'd do it in this way. 

Assume its true that it's true that $2^n&lt;n!=&gt;\frac{2^n}{n!}&lt;1$ then for $n+1$ we have to prove that:

$2^{n+1}&lt;(n+1)!$

$2^n&lt;n!(n+1) * 1/2$

$\frac{2^n}{n!}&lt;(n+1)*1/2$

And then state that as $\frac{2^n}{n!}&lt;1$ and $\frac{n+1}{2}&gt;1$ for any $n&gt;4$ Therefore the inequality is true.