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.