I am writing up some notes and the following occurred to me and I would like to see if there are **a variety of ways to prove it**. Just for reference, the identity pops out of equality between constant term evaluations of [Laurent polynomials from my earlier quest](https://mathoverflow.net/questions/449007/have-you-seen-this-constant-term-extraction). There should be easier ways as well, I think.



> 
> **QUESTION.** Can you provide a proof for
> $$\prod\_{j=0}^n\frac{(2j)!^2}{(n+j)!\,j!}=2^n.$$
> 
> 
> 



Product/quotient of factorials beget dyadic powers
This is easy. Start with $$n!.2^n=(2n)(2n-2)\dots 2.$$ This implies that $$((2n-1)!(2n-3)!\dots 1!).n!.2^n=(2n)!(2n-2)!\dots 2!.$$ Then $$((2n)!(2n-1)!\dots 1!).n!.2^n=((2n)!(2n-2)!\dots 2!)^2.$$ Thus $$\left(\prod\_{j=0}^n (n+j)!j!\right).2^n=\prod\_{j=0}^n (2j)!^2. $$ Finally, divide both sides by the product on the left.


