I am trying to determine if $n!$ is greater than $4^{n}$, one way to attempt to solve it was by using the method of induction, and I was told me in link the following solution.
Use induction on $n$. Prove the fact for some $n > 4$. Then $$4^{n+1} = 4\cdot 4^n < 4n! < (n+1)n! = (n+1)!$$
But how can $4\cdot 4^n$ be smaller than $4n!$ ?
if $n = 5$ so $4\cdot 4^5 = 4096$ and $4\cdot(5\cdot4\cdot3\cdot2\cdot1) = 480$
Could you help me understand this solution?
As pointed out in the comments it is stated that $n! > 4^n$ for some $n > 4$. It was also pointed out that the first $n$ for which the inequality holds is $n = 9$ since $9! = 362880 > 262144 = 4^9$. So we take some $n > 4$ to be $n = 9$ and set it as the base case and show the inequality holds for all $n \geq 9$ inductively.
Now, we assume $n! > 4^n$ and apply induction:
$$ (n + 1)! = (n + 1) \cdot n! > 4 \cdot n! > 4 \cdot 4^n = 4^{n + 1} $$ $$ (n + 1)! > 4^{n + 1} $$
We note the base case holds and so we conclude $n! > 4^n$ for all $n \geq 9$.
