As an off-shot of my earlier MO question, I have found a "really cute" identity. The connection is revealed in the limit $q\rightarrow 1$.

So, I would like to ask:

QUESTION. Is there a combinatorial (conceptual) proof of this equality? $$\prod_{i=1}^n\binom{2i}i^2=2^n\prod_{i=1}^n\binom{n+i}i\binom{n}i \qquad \text{or} \qquad \prod_{i=1}^n\binom{2i}i^2=2^n\prod_{i=1}^n\binom{n+i}{i,i,n-i}.$$

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    $\begingroup$ Note that $\binom{n+i}i\binom ni=\binom{n+i}{n-i}\binom{2i}i$. The two identities can be easily seen. $\endgroup$ Feb 6, 2022 at 16:45
  • $\begingroup$ This does not seem to address the question. Hope you agree. $\endgroup$ Feb 6, 2022 at 17:09
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    $\begingroup$ Perhaps you could say something about the proof you have, and what you find "non-conceptual" or "non-combinatorial" about it. I was able to verify the formula by brute force, breaking things down into factorials and using the "cute" identity $(2^n)(n!)((2n-1)!!) = (2n)!!$, but I don't really know if that qualifies as the sort of proof you're looking for. Also, to amplify Zhi-Wei Sun's comment, the formula he points out immediately shows that your formula is equivalent to $\prod \binom{2i}{i} =2^n \prod \binom{n+i}{n-i}$ -- maybe simpler? Is there a reason you've stated it this way instead? $\endgroup$
    – Tim Campion
    Feb 6, 2022 at 17:20
  • $\begingroup$ All that you (and Zhi-Wei) said makes sense, algebraically (and that is also how I could verify the identity). However, you can understand that such an approach would not be viewed as combinatorial (bijective proof, counting interpretations, etc). Hope this helps. $\endgroup$ Feb 6, 2022 at 17:24
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    $\begingroup$ @GHfromMO: removed. $\endgroup$ Feb 7, 2022 at 14:41


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