This "innocent-looking" identity came out of some calculation with determinants, and I like to inquire if one can provide a proof. Actually, different methods of proofs would be of valuable merit and instructional.



> 
> **Question.** Can you justify the following identity?
>  $$\prod\_{j=1}^n\binom{2j}j=\prod\_{j=1}^n2\binom{n+j}{2j}.$$
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> 



An identity for product of central binomials
We have
$$\prod\_{j=1}^n \binom{2j}{j} = \frac{2!4!\cdots (2n)!}{(1!2!\cdots n!)^2}$$
and
$$\prod\_{j=1}^n 2\binom{n+j}{2j} = 2^n\frac{(n+1)!(n+2)!\cdots (2n)!}{(2!4!\cdots (2n)!)\cdot (0!1!\cdots (n-1)!)} = \frac{(n+1)!(n+2)!\cdots (2n)!}{(1!3!\cdots (2n-1)!)\cdot (1!2!\cdots n!)}.$$
Diving the former by the latter, we get
$$\frac{(2!4!\cdots (2n)!)\cdot (1!3!\cdots (2n-1)!)}{(1!2!\cdots n!)\cdot ((n+1)!(n+2)!\cdots (2n)!)} = \frac{1!2!\cdots (2n)!}{1!2!\cdots (2n)!} = 1.$$


