<s>[INST] Proving $ { 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $ How to prove this  binomial identity  :
$$ { 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$
The left hand side arises while solving a standard binomial problem the right hand side is given in the solution , I checked using induction that this is true but I am inquisitive to prove it in a rather general way.
EDIT: I have gone through all of the answers posted here,I particularly liked Isaac♦
answers after which it was not much difficult for me to figure out something  i would rather say an easy and straight algebraic proof, I am posting it here if somebody needs in future :
$$ { 2n \choose n } = \frac{(2n)!}{n! \cdot n!} $$
$$ = \frac{ 1 \cdot 2 \cdot 3 \cdots n \cdot (n+1) \cdots (2n-1)\cdot (2n) }{n! \cdot n!}$$
$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdots (2n)]}{n! \cdot n!}  $$
$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [(2.1) \cdot (2.2) \cdot (2.3) \cdots (2.n)]}{n! \cdot n!}  $$
$$= \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [ 2^n \cdot (n)! ]}{n! \cdot n!}  $$
$$ = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$
I do always welcome your comments :-)
 [/INST] Your original identity, ${ 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!}$, can be rewritten (by multiplying both sides by $(n!)^2$) as $(2n)!=2^n\cdot 1\cdot 3\cdot 5\cdots (2n-1)\cdot n!$.  Now, $2^n\cdot 1\cdot 3\cdot 5\cdots (2n-1)\cdot n!=$ $1\cdot 3\cdot 5\cdots (2n-1)\cdot 2\cdot 4\cdot 6\cdots 2n=$ $1\cdot 2\cdot 3\cdots (2n-1)(2n)=(2n)!$.
</s>