$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$ Show that $$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$$
$$\sqrt{2}=\mathbf{2}^{1/2}$$
$$\sqrt{2\sqrt{2}}=\mathbf{2}^{1/2+1/2^2}$$
$$\sqrt{2\sqrt{2\sqrt{2}}}=\mathbf{2}^{1/2+1/2^2+1/2^3}$$
Show the limit of $$\mathbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+\dotsb+\frac{1}{2^n}=1$$ when $n\to\infty$
$$\textbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}$$
$$\Rightarrow \frac{1}{2}\textbf{S}_{n}=\frac{1}{2}(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n})$$
$$\Rightarrow \frac{1}{2}\textbf{S}_{n}=(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n+1}})$$
$$\Rightarrow (1)-(2)=\textbf{S}_{n}-\frac{1}{2}\textbf{S}_{n}=\frac{1}{2}-\frac{1}{2^{n+1}}$$
$$\Rightarrow \textbf{S}_{n}(1-\frac{1}{2})=\frac{1}{2}-\frac{1}{2^{n+1}}$$
$$\Rightarrow \frac{1}{2^{n+1}}\rightarrow\textbf{0}\quad\textit{when n}\rightarrow\infty$$
$$\Rightarrow \textbf{S}_{n}\rightarrow\textbf{1}\quad\textit{when n}\rightarrow\infty$$
$$\Rightarrow \lim_{n \to \infty}\textbf{2}^{\textbf{S}_{n}}=2\quad\textit{when n}\rightarrow\infty$$
 A: Let $x_1 = \sqrt{2}$ and define $x_{n+1} = \sqrt{2 x_n}$, then it suffices to show 
that $\lim_n x_n = 2$.  In order to achieve this goal show that the sequence
$x_n$ is monotonically increasing and bounded above (I will leave this for you to do).
Then the limit exists so let $x = \lim x_n$.  Then, using $x_{n+1} = \sqrt{2 x_n}$ and the continuity of the square root function, we get that $x = \sqrt{2 x} \implies x=2$.
A: Let S be your general term for a large value of n. If your square it you have S^2 = 2 S, then S = 2.
A: It's more simple to prove that, let $\sqrt{2\sqrt{2\sqrt{2}...}}=t$. Then, $t^2=2t \rightarrow t=2$.
A: $$T=\sqrt{2\sqrt{2\sqrt{2}}}...\\
\frac{T^2}{2}=T\\$$
T is a nonzero real number so:$$
T=2$$
A: Your proof is correct!
Here is a different proof. 
First note that the sequence, let's call it $(a_n)_{n\in\mathbb N}$, can be defined recursively as $a_1=\sqrt{2}$ and $a_{n+1}=\sqrt{2a_n}$. Then you can show that it is increasing and bounded above by 2, therefore it converges.
Its limit, $\ell$, must satisfy $\ell=\sqrt{2\ell}$, and therefore $\ell=2$.
