Show by mathematical induction that $(2n)! > 2^n*n!$ for all $n \geq 2$ Very stuck on this question. Here is what I have attempted.
Base case: $(2*2)! > 2^2*2!$ = $24>8$ Which is true
Hypothesis: Assume for some int k>2 that $(2k)!>2^k*k!$
Then for my inductive step, I am getting stuck. I tried,
$(2(k+1))! > 2^{k+1} *(k+1)!$
$(2k+2)!> 2^{k+1} *(k+1)! $
$(2k+2)(2k+1)(2k)!>2* 2^{k} *(k+1)! $
And then I'm stuck... I understand I am supposed to use my IH somewhere, but I do not understand how to effectively apply it in this step. Can somebody clearly show me how I should use the IH?

$$(2(k+1))! \overset{?}{>} 2^{k+1}(k+1)!$$
$$(2k+2)(2k+1)(2k)! \overset{?}{>} \cdot2^{k+1}(k+1)!$$
$$(2k+2)(2k+1)(2k)! \overset{?}{>} 2 \cdot2^k(k+1)k!$$
Now use your IH, since $(2k)! > 2\cdot2^k k!$, you just need to show that 
$$(2k+2)(2k+1) > (k+1),$$
which is obvious.
