Vieta's Formula for Pi

Theorem

 * $\pi = 2 \times \dfrac 2 {\sqrt 2} \times \dfrac 2 {\sqrt {2 + \sqrt 2} } \times \dfrac 2 {\sqrt {2 + \sqrt {2 + \sqrt 2} } } \times \dfrac 2 {\sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt 2 } } } } \times \cdots$

Proof
Thus:

Then we have from the Half Angle Formula for Cosine that:

So we can replace all the instances of $\cos \dfrac \pi 4$, $\cos \dfrac \pi 8$, etc. with their expansions in square roots of $2$.

Finally, we note that from Limit of Sine of X over X we have:
 * $\displaystyle \lim_{\theta \mathop \to 0} \frac {\sin \theta} {\theta} = 1$

As $n \to \infty$, then, we have that $\dfrac \pi {2^n} \to 0$, and so:
 * $\displaystyle \lim_{n \mathop \to \infty} \frac {\sin \left({\pi / 2^n}\right)} {\pi / 2^n} = 1$

The result follows after some algebra.