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'?

Note that $$I = \frac{n(n-1)(n-2)(n-3) \cdots 2 \cdot 1}{2n(2n-1)(2n-2)(2n-3) \cdots 2 \cdot 1} $$$$= \frac{n(n-1)(n-2)(n-3) \cdots 2 \cdot 1}{2^n \times n(2n-1)(n-1)(2n-3)(n-2) \cdots 1 \cdot 1}$$ and after cancelling out we get that$$ I = \frac{1}{2^n (2n-1) (2n-3) \cdots 3 \cdot 1} = \frac{1}{2^n} \prod_{i=1}^n \frac{1}{2i -1} $$This is one possible expression, but I think yours is more concise. 
