$$\eqalign{ & \mathop {\lim }\limits_{n \to \infty } {{n!} \over {(n + 1)(n + 2)...(2n)}} = {{n!} \over {{{(2n)!} \over {n!}}}} = \cr & {{n!n!} \over {2n!}} = {{n!} \over 2} = + \infty \cr} $$
The first equality can be explained by an example:
$$\eqalign{ & n = 5 \cr & {{5!} \over {6!...10!}} \cr} $$
Is it right? because something's telling me it isn't.
$n!$ can be expressed as a polynomial of $n$:
$$n! = \prod_{i=0}^n (n-i) = n^{n+1} + \cdots$$
This shows that $n!$ is a $n+1$-degree polynomial of $n$. Then, the greatest power of $n$ that you can find in this polynomial is $n^{n+1}$. The greatest power that you can obtain from $n!n!$ is $n^{n+1}n^{n+1} = n^{2n+2}$ with coefficient $1$. Also, the greatest power of the polynomial of $(2n)!$ is $(2n)^{2n+1}$ with coefficient 1.
The ratio between the maximum powers is : $$\frac{n^{2n+2}}{(2n)^{2n+1}} = \frac{n^{2n+1}n}{2^{2n+1}n^{2n+1}} = \frac{n}{2^{2n+1}}$$
This ratio determinates the behavior of the limit when $n$ goes to infinity. In particular:
$$\lim_{n\rightarrow +\infty} \frac{n!n!}{(2n)!} = \lim_{n\rightarrow +\infty}\frac{n}{2^{2n+1}} = 0$$