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?
  $$\prod_{j=1}^n\binom{2j}j=\prod_{j=1}^n2\binom{n+j}{2j}.$$

 A: 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
$$\frac{(2!4!\cdots (2n)!)\cdot (1!3!\cdots (2n-1)!)}{(1!2!\cdots n!)\cdot ((n+1)!(n+2)!\cdots (2n)!)} = \frac{1!2!\cdots (2n)!}{1!2!\cdots (2n)!} = 1.$$
A: Let me add my own approach, thereby offering a context and motivation for the identity at hand. Of course, it is not the easiest technique but that was not my goal or intent in asking the question above.
As I mentioned, things came out of determinantal consideration.
The story began with this MO question, in which I provided an alternative proof to the (well-known) special case of the $(n+1)\times(n+1)$ Hankel matrix
$$\det\left[\binom{2i+2j}{i+j}\right]_{i,j=0}^n=2^n.\tag 1$$
Method 1. Generalize (1) and prove that (check my argument here)
$$\det\left[\binom{2i+2j+2a}{i+j+a}\right]_{i,j=0}^n
=\prod_{j=0}^n\binom{2a+2j}{a+j}\binom{a+n+j}{2j}^{-1}.$$
In particular, setting $a=0$, the determinant (1) is computed by
$$\det\left[\binom{2i+2j}{i+j}\right]_{i,j=0}^n
=\prod_{j=1}^n\binom{2j}j\binom{n+j}{2j}^{-1}.\tag2$$
Method 2. Using the matrix decomposition (a product of triangular and diagonal matrices)
$$\left[\binom{2i+2j}{i+j}\right]_{i,j=0}^n
=\left[\binom{2i}{i-j}\right]\cdot D\cdot\left[\binom{2i}{i-j}\right]^T;\tag 3$$
where $D$ is the diagonal matrix $D=\text{diag}(1,2,2,\dots,2)$. Now, direct calculation in (3) generates
$$\det\left[\binom{2i+2j}{i+j}\right]_{i,j=0}^n=2^n. \tag4$$
Finally, equating (2) and (4) leads to the identity in our problem.
A: 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 $k$.
