Evaluating $\lim_{n \to \infty} \frac{(2n)!}{2^n (n!)^2}$ 
Can you please explain how should I evaluate this limit
$$\lim_{n \to \infty} \frac{(2n)!}{2^n (n!)^2}$$

I know the solution is $\geq1$ but I don't know how I can just simplify like this but I stuck here $$\frac{2n(2n-1)(2n-2)...1}{2^n(n(n-1)(n-2)...1)}$$ I don't know if I should simplify it like that as I said the answer is $\geq1$ but  my textbook doesn't explain why?

Let $a_n=\frac{(2n)!}{2^n (n!)^2}$ and consider the limit of the ratio
$$\frac{a_{n+1}}{a_n}=\frac{(2n+2)!}{2^{n+1} ((n+1)!)^2}\cdot \frac{2^n (n!)^2}{(2n)!}=\frac{(2n+2)(2n+1)}{2 (n+1)^2}\to \ ?$$
Do you know any theorem about limit of sequences which involves such ratio?
