Value of the product: $ \sqrt{2} \sqrt{2 - \sqrt{2}} \sqrt{2 - \sqrt{2 - \sqrt{2}}} \sqrt{2 - \sqrt{2 - \sqrt{2-\sqrt{2}}}} \cdots $ =? Let the recursive sequence
$$ a_0 = 0, \qquad a_{n+1} = \sqrt{2-a_n},\,\,n\in\mathbb N.
$$
T
Can we find the value of the product
$$
\prod_{n=1}^{\infty}{a_n}?
$$
Well, from here I don't seem to follow. I can understand that there would be some good simplification and the product will hopefully telescope but I'm lacking the right algebra. I also thought of finding a recurrence solution probably from the corresponding DE but that didn't follow as well.

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
$$
\sqrt{2}\sqrt{2-\sqrt{2}}\sqrt{2-\sqrt{2-\sqrt{2}}}\sqrt{(2-\sqrt{2-\sqrt{2-\sqrt{2}}}}\cdots = \sqrt{\frac{4}{3}}
$$
