<s>[INST] An identity for product of central binomials 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?
  $$\prod_{j=1}^n\binom{2j}j=\prod_{j=1}^n2\binom{n+j}{2j}.$$

 [/INST] I think that this is routine to verify by induction (as Fedor Petrov suggests in a comment). With respect to a response that is "instructional," suppose (or check) that the identity holds when $n = 4$. As one moves up to $n=5$, the right hand side changes from:
$$2\binom{5}{2}2\binom{6}{4}2\binom{7}{6}2\binom{8}{8}$$
to:
$$2\binom{6}{2}2\binom{7}{4}2\binom{8}{6}2\binom{9}{8}2\binom{10}{10} = 2\binom{6}{2}2\binom{7}{4}2\binom{8}{6}2\binom{9}{8}2$$
Using the fact that $\binom{n}{k} / \binom{n-1}{k} = \frac{n}{n-k}$, we find the ratio of the RHS for $n=5$ to the RHS when $n=4$ to be: 
$$(6/4)(7/3)(8/2)(9/1)2 = 2\frac{9!}{5!4!}$$
Meanwhile, the ratio of the LHS when $n=4$ to $n=5$ changes by a multiplicative factor of
$$\binom{10}{5} = \frac{10!}{5!5!} = \frac{10}{5}\frac{9!}{5!4!} = 2\frac{9!}{5!4!}$$
as above.
I believe the general proof simply requires a bit of attention paid to bookkeeping around $n$ and $k$.
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