# A (really!) cute identity between product of binomials

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? $$\prod_{i=1}^n\binom{2i}i^2=2^n\prod_{i=1}^n\binom{n+i}i\binom{n}i \qquad \text{or} \qquad \prod_{i=1}^n\binom{2i}i^2=2^n\prod_{i=1}^n\binom{n+i}{i,i,n-i}.$$

• Note that $\binom{n+i}i\binom ni=\binom{n+i}{n-i}\binom{2i}i$. The two identities can be easily seen. Feb 6, 2022 at 16:45
• This does not seem to address the question. Hope you agree. Feb 6, 2022 at 17:09
• 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? Feb 6, 2022 at 17:20
• 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. Feb 6, 2022 at 17:24
• @GHfromMO: removed. Feb 7, 2022 at 14:41