<s>[INST] How can we prove $\sqrt{2+\sqrt{2+....\sqrt{2+\sqrt{2}}}} = 2\cos\left(\frac {\pi }{2^{n+1}}\right)$ without induction I wanted to know the proof without induction / substitution method  for the equation
$$\underbrace {\sqrt{2+\sqrt{2+...\sqrt{2+\sqrt{2}}}} }_{\text{n-times}}= 2\cos\left(\frac {\pi }{2^{n+1}}\right)$$
Proof with induction:
\begin{align}
& n=1: \\
& \sqrt{2}=2\cos\left( \dfrac{\pi}{4} \right). \\
\ \\
& \text{Assume that the equation is valid when }n=k. \\
\ \\
&n=k+1; \\
& \underbrace {\sqrt{2+\sqrt{2+...\sqrt{2+\sqrt{2}}}} }_{\text{k+1-times}}=\sqrt{2+\underbrace {\sqrt{2+\sqrt{2+...\sqrt{2+\sqrt{2}}}} }_{\text{k-times}}} \\
& = \sqrt{2+2\cos\left( \dfrac{\pi}{2^{k+1}} \right)} = \sqrt{2}\cdot\sqrt{2\cos^2\left( \dfrac{\pi}{2^{k+2}} \right)} = 2\cos\left(\dfrac{\pi}{2^{k+2}}\right). \blacksquare
\end{align}
 [/INST] Observe that $2+\sqrt{2} = 2\left(1+\dfrac{1}{\sqrt{2}}\right)=2\left(1+\cos(\frac{\pi}{4})\right)= 2(2\cos^2\left(\frac{\pi}{8}\right))$. Thus when you take the innermost square root you get $2\cos\left(\frac{\pi}{8}\right)$. And repeat this $n$ times you will get to the formula you want to have.
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