Closed form for $\prod_{n=1}^\infty\sqrt{\tanh(2^n)},$

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.

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For $x < 1$, we have the Taylor series expansion:
$$f(x):= \frac{-1}{4} \log \left(- \frac{x – x^{-1}}{x + x^{-1}} \right) = \frac{x^2}{2} + \frac{x^6}{6} + \frac{x^{10}}{10} + \frac{x^{14}}{14} + \ldots $$

Then

$$f(x) + \frac{f(x^2)}{2} + \frac{f(x^4)}{4} + \frac{f(x^8)}{8} + \ldots
= \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{6} + \frac{x^8}{8} + \frac{x^{10}}{10} + \ldots $$
$$= – \frac{1}{2} \log(1 – x^2).$$

Now let $x = e^{-2}$. Then

$$ \log \left( \sqrt[2^n]{\mathrm{tanh}(2^n)} \right) =
\frac{1}{2^n} \log \left( \frac{e^{2^n} – e^{-2^n}}{e^{2^n} + e^{-2^n}}\right)
$$
$$= \frac{-4}{2^n} f(e^{-2^n}) = \frac{-4}{2^{n}} f(x^{2^{n-1}}),$$

Hence summing over all $n \ge 1$, we see that, if the product is $P$, then

$$\log P = -4 \sum_{n=0}^{\infty} \frac{1}{2^{n}} f(x^{2^{n-1}})
= -2 \sum_{n=1}^{\infty} \frac{1}{2^{n}} f(x^{2^{n}}) = \log(1 – x^2),$$

and thus

$$P = \exp \log(1 – x^2) = 1 – x^2 = 1 – e^{-4}.$$

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
$$
\prod_{n=1}^\infty\tanh(2^n)^{1/2^n}=1-e^{-4}\tag{8}
$$