Let
$A=2^{2^{-x}}$ and
$B=2^{2^{-x}+1}(1+2^{2^{-1}})(1+2^{2^{-2}})\cdots(1+2^{2^{-x+1}})$
Showing
$x\ge2$ $${\pi\over 2}=2\tan^{-1}\left({1\over A}\right)+\tan^{-1}\left({1\over B}\right)\tag1$$
Expanding $(1)$
Using $$\tan^{-1}\left({1\over A}\right)+\tan^{-1}\left({1\over B}\right)=tan^{-1}\left({A+B\over AB-1}\right)$$
$${\pi\over 2}=\tan^{-1}\left({A^2+2AB-1\over A^2B-2A-B}\right)\tag2$$
We know that $${\pi\over 2}=\lim_{y\to 0}\tan^{-1}\left({1 \over y}\right)$$
So this mean that if we only have to prove $A^2B-2A-B=0$
Am I in the right direction?
Is there another simple method where we can prove (1)?
Right direction. $A^2 B-2A -B=0$ is equivalent to: $$ \frac{2A}{B} = (A^2-1) \tag{1}$$ that in our case follow from the identity: $$ c^{2^n}-1 = \left(c^{2^{n-1}}+1\right)\cdot \left(c^{2^{n-2}}+1\right) \cdot \ldots \cdot\left(c^{2}+1\right)\cdot(c+1)\cdot(c-1).\tag{2} $$