How is it possible to prove that the multiplication from 0 to n of $$\cos{(x/2^k)}$$ with $x=\pi / 4$ is $$\dfrac{\sqrt{2}}{2} \times \dfrac{\sqrt{2+\sqrt{2}}}{2} \times \dfrac{\sqrt{2+\sqrt{2 + \sqrt{2}}}}{2} \times ...$$
I tried recurrence but got stuck
You can use the formula (valid for $0<x<\pi$) $$ \cos\Bigl(\frac{x}{2}\Bigr)=\sqrt{\frac{1+\cos x}{2}} $$ repeatedly. So, for the first terms, $$ \cos(\pi/4)=\frac{\sqrt{2}}{2},\quad \text{and}\quad \cos(\pi/8)=\sqrt{\frac{1+\cos(\pi/4)}{2}}=\sqrt{\frac{2+\sqrt{2}}{4}}=\frac{\sqrt{2+\sqrt{2}}}{2}. $$ Do you recognize the first two factors? Do you see the pattern?