Product of greatest common divisor

Let $n=x 2^y$ , where $x$ is odd. Then $\gcd\left(2^n,n \right)=2^y$

Then the product we need is $\begin{aligned} P&=\prod_{n=1}^\infty \gcd\left(2^n,n \right)^{\left(n^{-2}\right)} \\ &=\prod_{n=1}^\infty \left(2^y \right)^{\left(n^{-2}\right)} \\ &=\prod_{n=1}^\infty 2^\frac{y}{n^2} \\ &=\prod_{n=1}^\infty 2^\frac{y}{x^2 4^y} \\ &=\prod_{x \text{ odd}} \left[\prod_{y=0}^\infty 2^\frac{y}{x^2 4^y} \right] \\ &=\prod_{y=0}^\infty \left[\prod_{x \text{ odd}} 2^\frac{y}{x^2 4^y} \right] \\ &=\prod_{y=0}^\infty \left[\prod_{x \text{ odd}} 2^\frac{1}{x^2} \right]^\frac{y}{4^y} \end{aligned}$

(where some assumptions about convergence have been made). The bracketed quantity can be calculated as follows: $\prod_{x \text{ odd}} 2^\frac{1}{x^2} = 2^{\sum_{x \text{ odd}} \frac{1}{x^2}} = 2^\frac{\pi^2}{8}$

(using the fact the sum of the reciprocals of the odd squares is $\frac{\pi^2}{8}$ , which follows directly from $\zeta(2)=\frac{\pi^2}{6}$ ). So now $\begin{aligned} P&=\prod_{y=0}^\infty \left[2^\frac{\pi^2}{8}\right]^\frac{y}{4^y} \\ \log P &= \log \left[2^\frac{\pi^2}{8}\right] \sum_{y=0}^\infty \frac{y}{4^y} \\ &=\frac49 \log \left[2^\frac{\pi^2}{8}\right] \end{aligned}$

so finally $P=\left[2^\frac{\pi^2}{8}\right]^\frac49=2^\frac{\pi^2}{18} \approx \boxed{1.46}$

This seems to disagree slightly with the given answer, though.