A recursive sequence $c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$ We have a recursive sequence $$c_n =\sqrt{2+\sqrt{2+...+\sqrt{2}}}$$ with $n$ square roots.
We can obtain a recursive formula: $$c_n =\sqrt{2+c_{n-1}}\\
c_{n+1}=\sqrt{2+c_n}$$
Now we show that the sequence is increasing:
$$c_{n+1} \geq c_n \rightarrow\text{  This is what we need to prove}$$
$$ 
\sqrt{2+c_n} \geq \sqrt{2+c_{n-1}}\\
2+c_n \geq 2+c_{n-1}\\
c_n \geq c_{n-1} \implies c_{n+1} \geq c_n 
$$
Let's assume that a limit $c$ exists:
$$c = \sqrt{2+c}$$
$$c^2-c-2 = 0 \iff (c-2)(c-1)=0$$
So if the limit exists it is either 2 or 1. We know that it is not 1 since $c_1 > 1$ and the sequence is increasing.
$$c_n \leq 2\\
c_1 \text{ holds}\\$$
Now let's see for $n\rightarrow n+1$:
$$
c_{n+1} \leq 2\\
\sqrt{2+c_{n}} \leq \sqrt{2+2}\leq 2$$
Is the proof that the sequence is increasing  and is bounded sufficient?
 A: Your proof of growth is presented in a weird backward way. You could write
$$c_{n}\ge c_{n-1}\iff\sqrt{c_{n}+2}\ge\sqrt{c_{n-1}+2}\iff c_{n+1}\ge c_n.$$ You need a base case, which is missing. We indeed have $\sqrt{2+\sqrt2}\ge\sqrt2$.

You can also condense in a single, non-inductive proof of
$$0\le c_n\le2\implies c_n\le c_{n+1}\le2$$
or equivalently
$$0\le x\le2\implies x\le\sqrt{x+2}\le\sqrt{2+2}$$
because
$$x^2-x-2\le0$$ for $x\in[-1,2]$.
A: Assuming the nested radical is infinite we have $x=\sqrt{2+\sqrt{2+...+\sqrt{2}}}\Rightarrow x^2-2=x \iff x^2-x-2=0\Rightarrow x=2$
A: We you define the recursive formula you need to define $c_0 = \sqrt 2$
you don't need to define both $c_n$ and $c_{n+1}$.  Defining $c_n = \sqrt{2 + c_{n-1}}$ when $n>1$ is enough.  Or defining $c_{n+1}=\sqrt{2 + c_n}$ is enough.
We can streamline our prove that $c_n$ is increasing and bounded by $2$ in one fell induction step.
$0 \le \sqrt{2} \le c_0 < 2$.
And if $\sqrt {2} \le c_n < 2$ then $c_{n+1} =\sqrt {2 +c_n} > \sqrt {c_n^2 + c_n} > \sqrt{c_n^2} = c_n \ge \sqrt 2$.
And $c_{n+1}= \sqrt{2+c_n} < \sqrt{2 + 2} = 2$.
(Your inductions had no base cases!)
Then as we know it is bounded above and increasing $\lim_{n\to \infty} c_n=c$ exists.
And as $\lim_{n\to \infty} c_n = \lim_{n\to \infty} c_{n+1} = \lim_{n\to\infty} \sqrt {2 + c_n} = \sqrt{2 + \lim_{n\to\infty}{c_n}}$ so
$c = \sqrt {2+c} \ge 0$ so
$c^2 - c -2 =0$
$(c-2)(c+1) = 0$ so $c = 2$ or $c = -1$ but $c> 0$ so $c = 2$.
