Closed form for $\prod_{k=1}^n \binom{k^2+2k}{k^2}$

Question

Does anybody know how I can find a closed form for the expression $$ \prod_{k=1}^n \binom{k^2+2k}{k^2}? $$ There are many ways to handle such things, but with sum instead of product. Here, I have no idea how to start.

Answer 1

For every $k$, $$ \binom{k^2+2k}{k^2}=\frac{(k^2+2k)!}{(k^2)!\,(2k)!}=\frac{((k+1)^2)!}{(k^2)!}\,\frac1{(k+1)^2}\,\frac1{(2k)!}, $$ hence $$ \prod_{k=1}^n \binom{k^2+2k}{k^2}=((n+1)^2)!\,\left(\prod_{k=1}^n\frac1{k+1}\right)^2\,\left(\prod_{k=1}^n\frac1{(2k)!}\right)=\frac{((n+1)^2)!}{((n+1)!)^2}\prod_{k=1}^n\frac1{(2k)!}. $$ This reduces the problem to finding a "closed form" for the product $$ \prod_{k=1}^n(2k)!, $$ a task which does not seem doable if "closed form" has one of its usual meanings.

Answer 2

There is no closed form expression, but we can evaluate its asymptotics by first taking its logarithm, and then employing Stirling's approximation.