TITLE: prove that $\frac{(2n)!}{(n!)^2}$ is even if $n$ is a positive integer
QUESTION [38 upvotes]: Prove that $\frac{(2n)!}{(n!)^2}$ is even if $n$ is a positive integer. 

For clarity: the denominator is the only part being squared.

My thought process: 
The numerator is the product of the first $n$ even numbers and the product of the first $n$ odd numbers. 
That is, $(2n!) = (2n)(2n-2)(2n-4)\cdots(2n-1)(2n-3)(2n-5).$ In effect, the product of even numbers can be cancelled out with $n!$ resulting in the following quotient: 
$$ \frac{(2^n)(2n-1)(2n-3)}{(n!)}\;.$$ 
To me this looks even thanks to the powers of $2$. But I am not convinced for some reason. 
Was this approach to naive? 
Sorry for the poor notation, I don't know any coding languages, my apologies.

REPLY [9 votes]: An elegant answer:
$$\frac{(2n)!}{(n!)^2} = \dbinom{2n}n = \sum_{k=0}^n \dbinom{n}k \cdot \dbinom{n}{n-k} = \sum_{k=0}^n \dbinom{n}{k}^2 \equiv \sum_{k=0}^n \dbinom{n}k = 2^n \equiv 0 \pmod 2.$$
The main step is that if $m$ is an integer, $m^2 \equiv m \pmod 2$.