cancelling out of $\frac{n!}{2n!}$? Perhaps a rather strange question. I am doing an exercise and the last part ends: 
$I = \frac{n!}{(2n)!}$
Now I know that: 
$$I = \frac{n(n-1)(n-2)(n-3)\cdots}{2n(2n-1)(2n-2)(2n-3)\cdots }$$
Is it possible to cancel out some terms? Or to make it a little bit more 'friendly'?

Separating the even and odd terms, we have
$$ (2n)! = [ (2n)(2n-2)(2n-4)\dotsm \cdot 2 ] [(2n-1)(2n-3)\dotsm 1 ] \\
= 2^n [n(n-1)(n-2) \dotsm 1] [(2n-1)(2n-3)\dotsm 1] = 2^n n! (2n-1)!!, $$
where the double factorial is defined by $k!! = k(k-2)(k-4)\dotsm 3 \cdot 1 $ if $k$ is odd, or $\dotsm 4 \cdot 2$ if $k$ is even.
Alternatively, $(2n)!/n! = (2n)(2n-1)\dotsm (n+1) = (n+1)_{n}$, where $(a)_k$ is the rising factorial $ a(a+1)\dotsm (a+k-1) $.
