How to calculate the following product? How can we compute 
$$\prod_{i=n}^{1}\cos^{2}\left(\frac{2^{i}}{2^{n+1}}x\right)$$
for $0<x<\pi$ ?
Attempt:
$$
\begin{align*}
\prod_{i=n}^{1}\cos^{2}\left(\frac{2^{i}}{2^{n+1}}x\right) & =\prod_{i=n}^{1}\cos^{2}\left(\frac{2^{i}}{2^{n+1}}x\right)\\
 & =\prod_{i=n}^{1}\left(\frac{1}{2}+\frac{1}{2}\cos\left(\frac{2^{i}}{2^{n}}x\right)\right)\\
 & =\frac{1}{2^{n}}\prod_{i=n}^{1}\left(1+\frac{1}{2}\frac{\sin\left(2^{i+1-n}x\right)}{\sin\left(2^{i-n}x\right)}\right)\\
 & =?
\end{align*}
$$
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\mbox{Note that} &
\bbox[10px,#ffd]{\prod_{i = n}^{1}
\cos^{2}\pars{{2^{i} \over 2^{n + 1}}x}} =
\prod_{i = 1}^{n}
\cos^{2}\pars{{2^{i} \over 2^{n + 1}}x}
\\[5mm] & = 
\prod_{i = 1}^{n}\cos^{2}\pars{{2^{n + 1 - i} \over 2^{n + 1}}x}
=
\prod_{i = 1}^{n}\cos^{2}\pars{x \over 2^{i}}
\\[5mm] & =
\bracks{\prod_{i = 1}^{n}\cos\pars{x \over 2^{i}}}^{2}\label{1}\tag{1}
\end{align}

Moreover,
\begin{align}
\sin\pars{x} & = 2\sin\pars{x \over 2}\cos\pars{x \over 2} =
2^{2}\sin\pars{x \over 4}\cos\pars{x \over 4}\cos\pars{x \over 2}
\\[5mm] & =
2^{3}\sin\pars{x \over 2^{3}}\cos\pars{x \over 2^{3}}
\cos\pars{x \over 2^{2}}\cos\pars{x \over 2^{1}}
\\[5mm] & = \cdots =
2^{n}\sin\pars{x \over 2^{n}}\prod_{i = 1}^{n}\cos\pars{x \over 2^{i}}
\label{2}\tag{2}
\end{align}

\eqref{1} and \eqref{2} lead to

$$
\bbx{\bbox[10px,#ffd]{\prod_{i = n}^{1}\cos^{2}\pars{{2^{i} \over 2^{n + 1}}x}} =
{\sin^{2}\pars{x} \over 2^{2n}\,\sin^{2}\pars{x/2^{n}}}}
$$
