Simplification of $\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}}$ I'm having trouble understanding how this expression:
$$\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}} \cdot \left(\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}}{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\dots+\sqrt2}}}}}\right)=$$
got to this one:
$$\frac2{\sqrt2\cdot\sqrt{2+\sqrt2}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\dots}$$

Now that I am reading this correctly I think I can make some sense of the claim. 
Let $x=\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$ and $ y=\sqrt{2-\sqrt{2+\sqrt{2+\cdots}}}$ The "equation" you ask about is $$y=y\frac{x}{x}=\frac{2}{\sqrt{2}\cdot\sqrt{2+\sqrt{2}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\ \cdots} $$ Since the righthand side has a finite numerator and denominator an infinite product (of factors at least $\sqrt{2}$) this says $$y=0.$$ That does make sense, at least informally, since $x$ (if it makes any sense) is positive and satisfies $x=\sqrt{2+x}$ meaning $x^2=2+x$ and $x=2.$ Then, indeed, $$y=\sqrt{2-x}=\sqrt{0}=0. $$ But why that particular way of writing $0?$

The presumed meaning of the expression for $x$ is
$x=\lim_{n \to \infty}x_n$ where $x_0=0$, $x_1=\sqrt{2}$, $ x_2=\sqrt{2+\sqrt{2}}$ and in general $x_{n+1}=\sqrt{2+x_n}.$
Then we might also define 
$y=\lim_{n \to \infty}y_n$ where $y_1=\sqrt{2}$, $y_2=\sqrt{2-\sqrt{2}}$, $y_3=\sqrt{2-\sqrt{2+\sqrt{2}}}$ and in general $y_{n+1}=\sqrt{2-x_n}.$
So the thing to be explained is $$y=y\frac{x}{x}\stackrel{?}{=}\frac{2}{\prod_{j=1}^{\infty}x_j} .\tag{*}$$
The sense I make of this is as the limit (in a somewhat informal sense) of $$y_n=y_n\frac{x_n}{x_n}=\frac{2}{\prod_{j=1}^nx_j}. \tag{**}$$ This can be shown by induction.
The base case is $y_1=\frac{2}{x_1}$ i.e. $$\sqrt{2}=\sqrt{2}\frac{x_1}{x_1}=\sqrt{2}\frac{\sqrt{2}}{x_1}=\frac{2}{x_1} $$
then the needed induction step is $y_{n+1}=\frac{y_{n}}{x_{n+1}}$ and indeed:
$$y_{n+1}=y_{n+1}\frac{x_{n+1}}{x_{n+1}}=\sqrt{2-x_{n}}\frac{\sqrt{2+x_{n}}}{x_{n+1}}=\frac{\sqrt{4-x_{n}^2}}{x_{n+1}}=\frac{y_{n}}{x_{n+1}}.  $$
The last step makes use of $x_{n}^2=2+x_{n-1}$ so $\sqrt{4-x_{n}^2}=\sqrt{2-x_{n-1}}=y_{n}$

So, given $(**)$, I can make sense of $$y=\lim_{n \to \infty}y_n=\lim_{n \to \infty}y_n\frac{x_n}{x_n}=\lim_{n \to \infty}\frac{2}{\prod_{j=1}^nx_j}.$$ But, since my proof of $(**)$ is by induction, I do not see how one can, in one step, say $$y\frac{x}{x}=\frac{2}{\prod_{j=1}^{\infty}x_j} .$$

One way to more formally justify the value $x=2$ is to show $2-\frac{1}{2^{n-1}} \le x_n \lt 2$.
Finally, it does not directly relate to the specific question you ask, but see this question for a proof that $$y_n=2\sin{\frac{\pi}{2^{n+1}}}$$ so that
 $$ \frac{\pi}{2^n}-\frac{\left(\pi/2^n\right)^3}{24} \lt y_n \lt \frac{\pi}{2^n}                                                                                $$
