This is an infinite nested radical expression, so I need to find a way to simplify it. I notice that the radicand (the expression under the square root sign) is always 2 minus another square root of the same form. So I can let $x = \sqrt{2 -\!\sqrt{2 - \!\sqrt{2 - \!\sqrt{2 - \cdots}}}}$ and then square both sides to get $x^2 = 2 - x$. This is a quadratic equation that I can solve by rearranging and factoring. I get $x^2 + x - 2 = 0$, which factors as $(x + 2)(x - 1) = 0$. So the solutions are $x = -2$ and $x = 1$. However, I have to reject $x = -2$ because it is negative and the original expression was a square root, which is always non-negative. So the only valid solution is $x = 1$. This means that $\sqrt{2 -\!\sqrt{2 - \!\sqrt{2 - \!\sqrt{2 - \cdots}}}} = 1$.


Evaluate $\sqrt{2 -\!\sqrt{2 - \!\sqrt{2 - \!\sqrt{2 - \cdots}}}}$.