How can we compute
$$\prod_{i=n}^{1}\cos^{2}\left(\frac{2^{i}}{2^{n+1}}x\right)$$
for $0<x<\pi$ ?
Attempt:
$$ \begin{align*} \prod_{i=n}^{1}\cos^{2}\left(\frac{2^{i}}{2^{n+1}}x\right) & =\prod_{i=n}^{1}\cos^{2}\left(\frac{2^{i}}{2^{n+1}}x\right)\\ & =\prod_{i=n}^{1}\left(\frac{1}{2}+\frac{1}{2}\cos\left(\frac{2^{i}}{2^{n}}x\right)\right)\\ & =\frac{1}{2^{n}}\prod_{i=n}^{1}\left(1+\frac{1}{2}\frac{\sin\left(2^{i+1-n}x\right)}{\sin\left(2^{i-n}x\right)}\right)\\ & =? \end{align*} $$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}
$$ \bbx{\bbox[10px,#ffd]{\prod_{i = n}^{1}\cos^{2}\pars{{2^{i} \over 2^{n + 1}}x}} = {\sin^{2}\pars{x} \over 2^{2n}\,\sin^{2}\pars{x/2^{n}}}} $$