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 :-)
 A: If you have proved it via induction, the you have proved it.
I don't know what you mean by "in a rather general way".
But note that
$$1\cdot3\cdot5\cdot\cdots\cdot(2n-1)$$
is just crying out to be rewritten as
$$\frac{1\cdot2\cdot3\cdot\cdots\cdot(2n)}
{2\cdot4\cdot6\cdot\cdots\cdot(2n)}.$$
A: 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, $2^n\cdot 1\cdot 3\cdot 5\cdots (2n-1)\cdot n!=$ $1\cdot 3\cdot 5\cdots (2n-1)\cdot 2\cdot 4\cdot 6\cdots 2n=$ $1\cdot 2\cdot 3\cdots (2n-1)(2n)=(2n)!$.
A: Here's the more general way that you seek. The LHS and RHS are both products of rational functions of $\rm\:n\:$, i.e. they satisfy $\rm\ f(n+1) = r(n)\ f(n)\ $ for some rational function $\rm r(n)\:$. To test whether two such functions are equal you need only test that they have the same value of $\rm\ r(n) = f(n+1)/f(n)\ $ and the same initial condition $\rm\: f(0)$.$\ $ For the above it's trivial: $\rm\ f(0) = 1\:$ and the common value of $\rm\ r(n)\ $ is 
$\rm\quad\quad\quad\displaystyle f(n)\ \ =\  \ {2n \choose n}\ \: =\: \ \frac{(2n)!}{n!^2}\ \ \Rightarrow\ \frac{f(n+1)}{f(n)}\ =\ \frac{2\:(n+1)\:(2n+1)}{\ (n+1)\ (n+1)}$ 
$\rm\quad\quad\quad\displaystyle f(n)\ =\ 2^n \frac{ 1 \cdot 3  \cdots (2n-1)}{n!} \ \Rightarrow\ \frac{f(n+1)}{f(n)}\ =\ \frac{2\ (2n+1)}{n+1}$  
Notice that this method is completely algorithmic - it requires no insight whatsoever. In essence one is employing the uniqueness theorem for first order difference equations  (recurrences) to prove an equality by simply checking that both sides satisfy the same difference / differential equation and initial conditions. Such techniques generalize to much wider classes of multivariate mixed differential - difference equations, e.g.  holonomic functions. For some further examples of this powerful technique see my post here.
A: Multiply right side by $\frac{n!}{n!}$, and distribute the $2^n$.
A: Here's an enumerative proof. Consider the following procedure for dividing a set $S$ of size $2n$ into two sets $A,B$ of size $n$. The procedure proceeds in $n$ rounds. We start with $A,B$ empty. In each round, let $m = \min S$ and pick some $x \in S \setminus m$. Now decide whether to put $m$ in $A$ and $x$ in $B$, or to put $m$ in $B$ and $x$ in $A$. Remove both $m,x$ from $S$.
Clearly, the procedure ends with the required partition, and requires $(2n-1)(2)(2n-3)(2)\cdots(1)(2)$ choices. Furthermore, I claim that each partition can be generated in $n!$ ways. Indeed, at each round, either $m \in A$ or $m \in B$; suppose wlog that $m \in A$. This forces our second choice, and for our force choice, we can choose an arbitrary member of $B$. So our total degrees of freedom are $(n)(n-1)\cdots(1)$.
