How to prove $ (2n)! < {2}^{2n} (n!)^2 $

Question

I am proving this statement.

$$ (2n)! < {2}^{2n} (n!)^2 $$

I proved it for $n=1$.

Did the induction hypothesis for $n+1$.

$(2n+2)! < 2^{2n+2}((n+1)!)^2 $

But I get stuck at this step,

$(2n+2)(2n+1)(2n)! < 2^{n} * 2^2 (n+1)^2 (n!)^2 $

What should I do next?

Answer 1

$$\begin{align} 2^{2n}=(1+1)^{2n}=\sum_{r=0}^{2n}\binom {2n}r\; &> \; \binom {2n}n=\frac {(2n)!}{n!n!}.\\ \therefore\ (2n)!\; &<\; 2^{2n}(n!)^2.\end{align}$$

Answer 2

You have $$\frac {(2n)!}{(n!)^2}=\frac {2n\cdot2(n-1)\cdot2(n-2)\dots 2}{n\cdot(n-1)\cdot{n-2}\dots 1}\cdot\frac {2n-1}{n}\cdot\frac{2n-3}{n-1}\dots\cdot \frac 11$$

The first fraction is just $2^n$ (taking the even numbers from $(2n)!$) and each of the $n$ factors in the second part (odd numerators) is less than $2$

@SimplyBeautifulArt has answered your main question in a comment, so this is just an alternative way of looking at things.

Answer 3

Equivalently, you want $$2^{2n} > \binom{2n}{n}$$ which is combinatorially obvious: the $n$-sized subsets of $\{1,2,\dots,2n\}$ can't outnumber the total subsets of that set.