Limit $ \sqrt{2\sqrt{2\sqrt{2 \cdots}}}$ 
Find the limit of the sequence $$\left\{\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots\right\}$$

Another way to write this sequence is $$\left\{2^{\frac{1}{2}},\hspace{5 pt} 2^{\frac{1}{2}}2^{\frac{1}{4}},\hspace{5 pt}  2^{\frac{1}{2}}2^{\frac{1}{4}}2^{\frac{1}{8}},\hspace{5 pt}  \dots,\hspace{5 pt}2^{\frac{1}{2^{n+1} - 2}}\right\}$$
So basically we have to find $\lim_{n\to\infty}2^{\frac{1}{2^{n+1} - 2}}$. This equates to $$\lim_{n\to\infty}2^{\frac{1}{2^{\infty+1} - 2}} \Longrightarrow 2^{\frac{1}{2^{\infty}}} \Longrightarrow 1$$
Is this correct? P.S. Finding that $S(n)$ was a pain!

Hints:
$$x=\sqrt{2\sqrt{2\sqrt2\ldots}}\implies x=\sqrt{2x}\implies x^2=2x\implies\ldots$$
