How to prove this combinatorial equality? How to prove the following equality:
$$\sum_{m=0}^n\frac{(2m-1)!!}{(2m)!!}\cdot\frac{(2n-2m-1)!!}{(2n-2m)!!}=1$$
I have transformed the equality into 
$$\frac{{n \choose 2n}}{2^{2n}}\sum_{m=0}^n\frac{{m \choose n}^2}{{2m \choose 2n}}=1$$
But I still don't know how to prove the equality.

By applying the general binomial theorem, we can show that
$$(1-x)^{-1/2} = \sum_{n=0}^\infty \binom{-1/2}{n} (-x)^n = \sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!}x^n$$
By squaring both sides, we have
$$\sum_{n=0}^\infty 1 \cdot x^n = (1-x)^{-1} = \sum_{n=0}^\infty \left(\sum_{m=0}^n\frac{(2m-1)!!}{(2m)!!}\frac{(2(n-m)-1)!!}{(2(n-m))!!}\right)x^n$$
from which the result follows by comparing coefficients.
