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}

Let's assume the problem. Then, let $x_n=2\cos \left( \dfrac {\pi}{2^{n+1}} \right).$ We can easily get $x_{n+1}=\sqrt{2+x_n}, x_1=\sqrt{2}$. Then, we can change the problem:

Prove that if $x_{n+1}=\sqrt{2+x_n}, x_1=\sqrt{2}$, then $x_n=2\cos \left( \dfrac {\pi} {2^{n+1}} \right)$.

We also know that $\displaystyle \lim_{n \to \infty}x_n=2$, because $x_n=\sqrt{2+x_n}.$ Also, $x_n$ is an increasing sequence. So, we have that $0<x_i\leq2$ for all $i$.
So, let $x_i=2\cos(a_i).(0<a_i<\frac{\pi}{2}.)$ Then, we have to prove that $a_n=\dfrac{\pi}{2^{n+1}}$.
$2\cos(a_{n+1})=x_{n+1}=\sqrt{2+x_n}=\sqrt{2+2\cos(a_n)}=\sqrt{2(1+\cos(a_n)}=\sqrt{2 \cdot 2\cos^2\left(\frac{a_n}{2}\right)}=2\cos\left( \frac{a_n}{2} \right).$
Since $0<a_i<\frac {\pi}{2}$, $\cos$ is injective. Therefore, $a_{n+1}=\frac{a_n}{2}$.
Then, we get:
$$a_{n+1}=\dfrac{a_1}{2^n}.$$
$x_1=\sqrt{2}=2\cos\left( \frac {\pi}{4} \right), a_1=\frac{\pi}{4}.$
Finally, we get:
$$a_n=\dfrac{\pi}{2^{n+1}.}$$
