$n,i\in\mathbb N$.


The summation in question is


$$\sum\_{k=1}^n\prod\_{l=1}^k\binom{2^n}{2^l}^i.$$


How fast does this grow? I am specifically looking at $i=1,2$.



How fast does this summation grow?
Let us obtain the asymptotic of $\log S\_n$, where $\log$ denotes the base-$2$ logarithm,
\begin{equation\*}
 S\_n:=\sum\_{k=1}^n\prod\_{l=1}^k\binom{2^n}{2^l}^i
 =\sum\_{k=1}^n a\_{n,k}, 
\end{equation\*}
and
\begin{equation\*}
 a\_{n,k}:=\prod\_{l=1}^k\binom{2^n}{2^l}^i. 
\end{equation\*}


For $k=1,\dots,n-1$,
\begin{equation\*}
 r\_k:=\frac{a\_{n,k+1}}{a\_{n,k}}=\binom{2^n}{2^{k+1}}^i, 
\end{equation\*}
so that $r\_{n-1}=1$ and
\begin{equation\*}
 r\_k\ge\binom{2^n}{2^{1+1}}^i\to\infty 
\end{equation\*}
(as $n\to\infty$) for $k=1,\dots,n-2$. So, $S\_n\sim a\_{n,n}+a\_{n,n-1}=2a\_{n,n}$ and
\begin{equation\*}
 \log S\_n=\log a\_{n,n}+O(1). \tag{1}\label{1}
\end{equation\*}
Next,
\begin{equation\*}
 \log a\_{n,n}=i\sum\_{l=1}^n L\_{n,l}=i\sum\_{l=1}^{n-1} L\_{n,l}, \tag{2}\label{2}
\end{equation\*}
where
\begin{equation\*}
 L\_{n,l}:=\log\binom{2^n}{2^l}. 
\end{equation\*}
By Stirling's formula, $\log N!=N\log(N/e)+O(\log(N+1))$ for integers $N\ge1$. So, for $l=1,\dots,n-1$,
\begin{equation\*}
 L\_{n,l}=2^n K\_{n,l}+O(n), \tag{3}\label{3}
\end{equation\*}
where
\begin{equation\*}
 K\_{n,l}:=K\_{n,l;1}-K\_{n,l;2}+K\_{n,l;3}, \tag{4}\label{4}
\end{equation\*}
\begin{equation\*}
 K\_{n,l;1}:=2^{l-n}(n-l),\quad K\_{n,l;2}:=\log(1-2^{l-n}),\quad K\_{n,l;3}:=2^{l-n}\log(1-2^{l-n}).
\end{equation\*}
Further,
\begin{equation\*}
 \sum\_{l=1}^{n-1} K\_{n,l;1}=2 (1 - (n+1)2^{-n})\to2; \tag{5}\label{5}
\end{equation\*}
\begin{equation\*}
 \sum\_{l=1}^{n-1} K\_{n,l;2}=\sum\_{l=-\infty}^{n-1} K\_{n,l;2}-\sum\_{l=-\infty}^0 K\_{n,l;2}
 =-c\_1+O(2^{-n})\to-c\_1, \tag{6}\label{6}
\end{equation\*}
where $c\_1:=-\sum\_{l=-\infty}^{n-1} K\_{n,l;2}=-\sum\_{j=1}^\infty\log(1-2^{-j})=1.7919\dots$;
\begin{equation\*}
 \sum\_{l=1}^{n-1} K\_{n,l;3}=\sum\_{l=-\infty}^{n-1} K\_{n,l;3}-\sum\_{l=-\infty}^0 K\_{n,l;3}
 =-c\_2+O(2^{-2n})\to-c\_2, \tag{7}\label{7}
\end{equation\*}
where $c\_2:=-\sum\_{l=-\infty}^{n-1} K\_{n,l;3}=-\sum\_{j=1}^\infty 2^{-j}\log(1-2^{-j})=0.63556\dots$.


Collecting \eqref{1}--\eqref{7}, we get
\begin{equation\*}
 \log S\_n\sim c\,i\,2^n,
\end{equation\*}
where $c:=2+c\_1-c\_2=3.1563\dots$.


