Kirsten Bishop

2022-11-29

Showing that if n is a natural number larger than 3, then $n!>{2}^{n}$
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 $P\left(k\right)$ is true.
$k!>{2}^{k}$
Now we need to show that $P\left(k+1\right)$ is true.
$\left(k+1\right)!={2}^{k+1}$
Proof:
$\left(k+1\right)!>\left(k+1\right)k!$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(k+1\right){2}^{k}$
After this I have no idea how to solve further.
Can anyone explain how to continue.

Gwendolyn Case

Expert

After finding out that the base case is true and assuming $P\left(k\right)$ is true, for $P\left(k+1\right)$ we have
by inductive argument and since $k+1>2$ we have $\left(k+1\right)!>{2}^{k+1}$

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