So, first step, base of induction is correct. When we take $n=9$, we get
$262114\leq 362880$.
So, we assume that
$2^{2n}\leq n!$ for all $n\geq 9$ is correct and we want to prove that
$2^{2n+2}\leq (n+1)!$ is correct for all $n\geq 9$.
In induction part, we take $2^{2n}\leq n!$ and multiply it by 4, so on the left side we get $2^{2n+2}$, what is left on the right side is $4n!$.
Now we need to prove that $2^{2n+2}\leq (n+1)!$ is correct.
Can we do that by checking if $(n+1)! \geq 4n!$, because we know for sure that $4n! \geq 2^{2n+2}$ ?
By induction, you take $n=9$ and get $262114\leq 362880$.
Now, assume $2^{2n}\leq n!$ is correct for all numbers up to some $n>9$.
$$(n+1)!=(n+1)\cdot n!\geq (n+1)\cdot 2^{2n}\geq 4\cdot 2^{2n}=2^{2(n+1)},\;\forall n>9,$$
where the induction hypothesis has been used in the first "$\;\geq$". Thus,
$$2^{2n}\leq n!,\;\forall n\geq 9.$$