The following is an induction argument, but the way it is written out, it doesn't look like it. We are looking at $\frac{(1)(2)(3)(4)\cdots(2n)}{n!}$ Separate out the odd terms on top. We get $\frac{[(1)(3)(5)\cdots(2n-1)][(2)(4)(6)\cdots (2n)]}{n!}.$
But $(2)(4)(6)\cdots(2n)=2^n n!$. So our quotient is $2^n(1)(3)(5)\cdots(2n-1).$ We conclude that $2^n$ divides our original expression. Not only that, it is the highest power of $2$ that does.
The idea can be turned into a textbook style induction. We prove that the highest power of $2$ that divides $P(2n,n)$ is $2^n$. The result is clearly true at $n=1$. Suppose it is true at $n=k$. We show that it is true at $n=k+1$. Note that $P(2k,k)=(2k)(2k-1)\cdots (k+2)(k+1),$ and $P(2k+2,k+1)=(2k+2)(2k+1)(2k)\cdots (k+2).$ Divide, noting that most of the terms cancel. We get $\frac{P(2k+2, k+1)}{P(2k,k)}=\frac{(2k+2)(2k+1)}{k+1}=2(2k+1),$ and therefore $P(2k+2,k+1)=2(2k+1)P(2k,k).$ Thus if $2^k$ is the highest power of $2$ that divides $P(2k,k)$, then $2^{k+1}$ is the highest power of $2$ that divides $P(2k+2,k+1)$.
Comment: Induction is exceedingly common, and usually not noticed. Roughly speaking, when we manipulate an expression that involves $\dots$, there is induction going on,