Asymptotic growth of $a(n)=\prod_{k\in\{1,\dots,n\}}{2^{k+1} \choose 2^k}$

Question

What's the asymptotic growth of:

$$ a(n)=\prod_{k\in\{1,\dots,n\}}{2^{k+1} \choose 2^k}\quad?$$

So far I was able to find only some lower bounds. For example, $a(n)>\prod_{k\in \{1\dots n\}} 2^{k+1}=\Theta(2^n)$

Also, I was thinking that it would be easy to find it in OEIS (and there to find some data about it, e.g. its asymptotic growth), but unfortuantely I could find it there.

Answer 1

We have $\binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}}$, hence by considering $n=2^k$ and multiplying both sides for $k\in\{1,2,\ldots,N\}$ we have that

$$\prod_{k=1}^{N}\binom{2^{k+1}}{2^k}\sim \prod_{k=1}^{N}\frac{4^{2^k}}{\sqrt{\pi}\,2^{k/2}}=\frac{4^{2^{N+1}-2}}{\pi^{N/2}2^{N(N+1)/4}} $$ which essentially is $2^{2^{N+2}}$.