How to prove $\binom{2n}{n}\frac{1}{n+1} = \prod \limits_{i = 2}^n \frac{2i-1}{i+1} $? How to prove this closed form involving Catalan numbers? $$\binom{2n}{n}\frac{1}{n+1} = \prod \limits_{i = 2}^n \frac{2 \times (2i-1)}{i+1} $$
I have seen this being used here. Not sure how to derive it. Any ideas?

Note that $\binom{2n}{n}=\frac{(2n)!}{n!n!}$. But 
$$(2n)!=\left(1\cdot 3\cdot 5\cdots \cdot(2n-1)\right)\left(2\cdot 4\cdot 6\cdots\cdot(2n)\right)$$
(we separated the odd numbers and the even numbers).  
The product of the even numbers $2\cdot 4\cdot 6\cdots(2n)$ is $2^n n!$.
From this it follows that
$$\frac{1}{n+1}\binom{2n}{n}=\frac{2^n}{(n+1)!}\left(1\cdot 3\cdot5 \cdots \cdot(2n-1)\right).$$
That is equal to the expression of the OP. For it can be written as 
$$\prod_{i=1}^n \frac{2\cdot(2i-1)}{i+1}.$$
Starting at $i=2$, as the revised OP does, eliminates one factor of $2$ from both the numerator and the denominator, so makes no difference. 
