The idea is to "complete the factorials":
$ 1\cdot 3 \cdot 5 \cdots (2n-1) = \frac{ 1 \cdot 2 \cdot 3 \cdot 4 \cdots (2n-1)\cdot (2n) }{2\cdot 4 \cdot 6 \cdots (2n)} $
Now take out the factor of $2$ from each term in the denominator:
$ = \frac{ (2n)! }{2^n \left( 1\cdot 2 \cdot 3 \cdots n \right)} = \frac{(2n)!}{2^n n!}$
A mathematician may object that there is a small gray area about what exactly happens between those ellipses, so for a completely rigorous proof one would take my post and incorporate it into a proof by induction.