Closed form for $\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$ Please help me to find a closed form for the infinite product
$$\prod_{n=1}^\infty\sqrt[2^n]{\tanh(2^n)},$$
where $\tanh(z)=\frac{e^z-e^{-z}}{e^z+e^{-z}}$  is the hyperbolic tangent.

Let
$$
f(x)=\prod_{n=0}^\infty\left(1-x^{2^n}\right)^{1/2^n}\tag{1}
$$
and
$$
g(x)=\prod_{n=0}^\infty\left(1+x^{2^n}\right)^{1/2^n}\tag{2}
$$
Then
$$
\begin{align}
f(x)\,g(x)
&=\prod_{n=0}^\infty\left(1-x^{2^{n+1}}\right)^{1/2^n}\\
&=\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{2/2^n}\\
&=\left(\frac{f(x)}{1-x}\right)^2\tag{3}
\end{align}
$$
from which we get
$$
\frac{f(x)}{g(x)}=(1-x)^2\tag{4}
$$
Note that
$$
\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{1/2^n}=\frac{f(x)}{1-x}\tag{5}
$$
and
$$
\prod_{n=1}^\infty\left(1+x^{2^n}\right)^{1/2^n}=\frac{g(x)}{1+x}\tag{6}
$$
Therefore, combining $(4)$, $(5)$, and $(6)$, we get
$$
\frac{\displaystyle\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{1/2^n}}{\displaystyle\prod_{n=1}^\infty\left(1+x^{2^n}\right)^{1/2^n}}=1-x^2\tag{7}
$$
Plug $x=e^{-2}$ into $(7)$ to get
$$
\prod_{n=1}^\infty\tanh(2^n)^{1/2^n}=1-e^{-4}\tag{8}
$$
