Convergence of sequence: $ \sqrt{2} \sqrt{2 - \sqrt{2}} \sqrt{2 - \sqrt{2 - \sqrt{2}}} \sqrt{2 - \sqrt{2 - \sqrt{2-\sqrt{2}}}} \cdots $ =?

Squaring the infinite product we observe that $$ 2(2-\sqrt{2})\big(2-\sqrt{2-\sqrt{2}}\big)\Big(2-\sqrt{2-\sqrt{2-\sqrt{2}}}\Big)\cdots\\ = 2\cdot\frac{2}{2+\sqrt{2}}\cdot\frac{2+\sqrt{2}}{2+\sqrt{2-\sqrt{2}}} \cdot\frac{2+\sqrt{2-\sqrt{2}}}{2+\sqrt{2-\sqrt{2-\sqrt{2}}}}\cdots \\ =\frac{4}{2+\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}} $$ But $$ \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\cdots}}}}= 1. $$ See: Convergence of $a_{n+1}=\sqrt{2-a_n}$. Thus $$ 2(2-\sqrt{2})\big(2-\sqrt{2-\sqrt{2}}\big)\Big(2-\sqrt{2-\sqrt{2-\sqrt{2}}}\Big)\cdots = \frac{4}{3} $$ Finally $$ \sqrt{2}\sqrt{2-\sqrt{2}}\sqrt{2-\sqrt{2-\sqrt{2}}}\sqrt{(2-\sqrt{2-\sqrt{2-\sqrt{2}}}}\cdots = \sqrt{\frac{4}{3}} $$

Another approach is the following one: if we assume $ a_n = 2\cos(\theta_n) $ it follows that $$ \cos(\theta_{n+1})=\sqrt{\frac{1-\cos\theta_n}{2}} = \sin\left(\frac{\theta_n}{2}\right) = \cos\left(\frac{\pi-\theta_n}{2}\right)\tag{1} $$ from which we have $\theta_{n+1}=\frac{\pi-\theta_n}{2}$ and, by induction: $$ \theta_{n+k} = \frac{\pi}{3}-(-1)^k\frac{\pi}{3\cdot 2^k}+(-1)^k\frac{\theta_n}{2^k}.\tag{2}$$ Since $\theta_0=\frac{\pi}{2}$, $$ \theta_k = \frac{\pi}{3}+(-1)^k \frac{\pi}{6\cdot 2^k},\qquad \color{red}{a_k = \cos\left(\frac{\pi}{6\cdot 2^k}\right)-(-1)^k\sqrt{3}\sin\left(\frac{\pi}{6\cdot 2^k}\right)}\tag{3} $$ but since $2\cos\theta_n = \frac{\sin(2\theta_n)}{\sin(\theta_n)}$ and $\sin(\pi-\theta)=\sin(\theta)$, we also have a telescopic product.

In particular: $$ a_1\cdot a_2\cdot\ldots\cdot a_n = \frac{\sin(2\theta_1)}{\sin(\theta_n)} \tag{4}$$ hence:

$$ \prod_{n\geq 1} a_n = \frac{\sin(2\theta_1)}{\sin(\lim_{n\to +\infty}\theta_n)} = \frac{1}{\sin\frac{\pi}{3}}=\color{red}{\frac{2}{\sqrt{3}}}.\tag{5}$$