trigonometry - Proof of an equality involving cosine $sqrt{2 + sqrt{2 + cdots + sqrt{2 + sqrt{2}}}} = 2cos (pi/2^{n+1})$

so I stumbled upon this equation/formula, and I have no idea how to prove it. I don't know how should I approach it:
$$
\sqrt{2 + \sqrt{2 + \cdots + \sqrt{2 + \sqrt{\vphantom{\large A}2\,}\,}\,}\,}\
=\
2\cos\left(\vphantom{\Large A}\pi \over 2^{n + 1}\right)
$$

where $n\in\mathbb N$ and the square root sign appears $n$-times.

I thought about using sequences and limits, to express the LHS as a recurrence relation but I didn't get anywhere.

edit: Solved, thanks for your answers and comments.

Answer

Hint:

Use induction and the half-angle formula for cosine.

Solution:

For $n=1$, the claim is true, since $\cos(\pi/4)=\sqrt{2}/2$. By the half-angle formula $$2\cos(x/2)=\sqrt{2+2\cos(x)}$$
Therefore
$$\sqrt{2+\sqrt{2+\cdots+\sqrt{2+\sqrt{2}}}}=\sqrt{2+2\cos\left(\frac{\pi}{2^n}\right)}=2\cos\left(\frac{\pi}{2^{n+1}}\right)$$
where in the left square root expressions there are $n$ square roots and in the first equality we have used the induction hypothesis that the claim holds for $n-1$.