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?

You're almost there:
$$
\begin{align}
(2(k+1))! &= (2k+2)(2k+1)(2k)! \\
&> (2k+2)(2k+1) 2^k k! \quad \text{(by the induction hypothesis)}\\
&= 2(k+1)(2k+1) 2^k k! \\
&= (2k+1) 2^{k+1} (k+1)! \\
&> 2^{k+1} (k+1)!
\end{align}
$$
Without induction:
$$
(2n)! = (1 \cdot 2 \cdots n) \cdot ((n+1) \cdots (2n)) > n! \cdot (2 \cdots 2) = 2^n \, n!
$$
