Showing that if n is a natural number larger than $3$, then $n! > 2^n$.
My try:
Base Case:
If $n=4$, then $4!>2^4$
$24>16$
So, the base case is true.
Assuming $P(k)$ is true.
$k!>2^k$
Now we need to show that $P(k+1)$ is true.
$(k+1)!=2^{k+1}$
Proof:
$(k+1)!>(k+1)k!$
$\implies (k+1)2^k$
After this I have no idea how to solve further.
Can anyone explain how to continue.
On
More generally, for any $m > 0$, if $n \ge em$ then $n! > m^n$.
Proof:
Start with $n! > (n/e)^n$. This can be proved by induction from $(1+1/n)^n < e$; this, in turn, has been shown here many times.
If $n \ge em$ then $n! > (n/e)^n \ge (em/e)^n =m^n$.
For your case with $m=2$, if $n \ge em \approx 5.4...$ or $n \ge 6$, $n! > 2^n$.
The remaining cases can be checked by hand.
You can also show, using $n! < (n/2)^n$ for $n \ge 6$, that if $m \ge 6$ and $n \le 2m$ then $n! < m^n$.
Therefore, if $r(n) = \dfrac{(n!)^{1/n}}{n}$, then $\dfrac1{e} \lt r(n) \lt \dfrac1{2} $ for $n \ge 6$.
After finding out that the base case is true and assuming $P(k)$ is true, for $P(k+1)$ we have
$$(k+1)! = (k+1)k! > (k+1)2^k$$
by inductive argument and since $k+1 > 2$ we have $(k+1)! > 2^{k+1}$.