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?
 A: $$(2(k+1))! \overset{?}{>} 2^{k+1}(k+1)!$$
$$(2k+2)(2k+1)(2k)! \overset{?}{>} \cdot2^{k+1}(k+1)!$$
$$(2k+2)(2k+1)(2k)! \overset{?}{>} 2 \cdot2^k(k+1)k!$$
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.
A: 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:
$$
(2n)! = (1 \cdot 2 \cdots n) \cdot ((n+1) \cdots (2n)) > n! \cdot (2 \cdots 2) = 2^n \, n!
$$
A: We need to show $$(2k+2)(2k+1)(2k)!>2* 2^{k} *(k+1)!$$
We know that $$(2k)!>2^k*k!$$
Thus $$(2k+2)(2k+1)(2k)!> (2k+2)(2k+1)2^k*k!>2^{k+1}*(k+1)!  $$
A: Using induction, you get
$$(2k+2)(2k+1)(2k)! = 2(k+1)(2k+1)(2k)! >$$
$$ > 2 \cdot 2^k \cdot (k+1)(k)!$$
Dividing both sides by $2(k+1)$ you are left with
$$(2k+1)(2k)! > 2^k(k)!$$
Using that $2k+1 \geq 1$ it the follows from your induction hypothesis.
