Proof of identity for $\pi$: $\frac{\pi}{3} = \frac{2}{\sqrt{2+\sqrt{3}}}\frac{2}{\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdots$

The nested square roots suggest Viète's formula and one way that formula can be derived, Euler's product identity for the sinc function: $$\frac{\sin x}x=\cos\frac x2\cos\frac x4\cos\frac x8\dots$$ Viète's formula is obtained by substituting $x=\frac\pi2$. We use $x=\frac\pi3$ instead: $$\frac{\sin\pi/3}{\pi/3}=\cos\frac\pi6\cos\frac\pi{12}\cos\frac\pi{24}\dots$$ $$\frac{\sqrt3/2}{\pi/3}=\frac{\sqrt3}2\cos\frac\pi{12}\cos\frac\pi{24}\dots$$ $$\frac3\pi=\cos\frac\pi{12}\cos\frac\pi{24}\dots$$ By applying the half-angle formula $\cos\frac x2=\sqrt{\frac{1+\cos x}2}=\frac{\sqrt{2+2\cos x}}2$ repeatedly we get $$\cos\frac\pi{12}=\frac{\sqrt{2+\sqrt3}}2$$ $$\cos\frac\pi{24}=\frac{\sqrt{2+\sqrt{2+\sqrt3}}}2$$ and so on, yielding $$\frac3\pi=\frac{\sqrt{2+\sqrt3}}2\cdot\frac{\sqrt{2+\sqrt{2+\sqrt3}}}2\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt3}}}}2\cdots$$ and from there the desired expression for $\pi$.