Why is this series of square root of twos equal $\pi$? Wikipedia claims this but only cites an offline proof:
$$\lim_{n\to\infty} 2^n \sqrt{2-\sqrt{2+\cdots+ \sqrt 2}} = \pi$$
for $n$ square roots and one minus sign. The formula is not the "usual" one, like Taylor series or something like that, so I can't easily prove it. I wrote a little script to calculate it and it's clearly visible, but that's not a proof.
 A: Here is a slightly different way to see why it is the area of a $2^k$-gon.  (It is really the same, I just also want to point out that the nested radical expression is the sin of $\frac{\pi}{2^k}$. This type of argument gives Vietas product formula for $\frac{2}{\pi}$)
Recall: If $a,b$ are sides of a triangle, and $\theta$ is the angle between them, then the area of this triangle is $\frac{1}{2}ab\sin(\theta)$.
For any regular $n$-gon inscribed in the unit circle, this means that by splitting it into the $n$ identical isosceles triangles with two sides equal to $1$, we have that $$\text{Area of regular n-gon}=\frac{n}{2}\cdot \sin\left(\frac{2\pi}{n}\right).$$
In the limit, this must approach $\pi$.  Now, the nested radical expression can be rewritten as sin of angle as follows:

We have that $$\sin\left(\frac{\pi}{2^k}\right)=\frac{1}{2}\sqrt{2-\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}.$$

Proof:  Notice that $$\cos(x)=\sqrt{\frac{1}{2}\left(1+\cos(2x)\right)}.$$  Using this iteratively allows us to find $\cos\left(\frac{\pi}{2^k}\right)$.  For example, since $\cos\left(\frac{\pi}{2}\right)=0$ we see that $$\cos\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}=\frac{1}{2}\cdot \sqrt{2}$$ and similarly
$$\cos\left(\frac{\pi}{8}\right)=\sqrt{\frac{1}{2}\left(1+\frac{1}{\sqrt{2}}\right)}=\frac{1}{2}\cdot \sqrt{2+\sqrt{2}}.$$  Iterating again we have $$\cos\left(\frac{\pi}{16}\right)=\frac{1}{2}\cdot \sqrt{2+\sqrt{2+\sqrt{2}}},$$ and by induction we'll arrive at our result.
Since $\cos^2(x)+\sin^2(x)=1$, one final manipulation will allow us to conclude our result. 
