How I can prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}$ converges to 2? Prove that the sequence $\sqrt{2} , \sqrt{2\sqrt{2}},  \sqrt{2\sqrt{2\sqrt{2}}} \ $ converges to $2$.


My attempt

I proved that the sequence is increasing and bounded by $2$, can anyone help me show that the sequence converges to $2$?
Thanks for your help.
We have 

$$\sqrt{2}=2^{\frac{1}{2}}.$$
Exciting, no?
We also have
$$\sqrt{2\sqrt{2}}=2^{\frac{1}{2}+\frac{1}{4}},$$
and 
$$\sqrt{2\sqrt{2\sqrt{2}}}=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}.$$
For the next term, we take the previous term, multiply by $2$, getting exponent of $2$ equal to $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}$, then take the square root, getting exponent of $2$ equal to $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}$.
The pattern continues, since the "next term" is always obtained by the same process.

It is well-known that the series $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$ converges to $1$. Since the function $f(x)=2^x$ is continuous, our original sequence converges to $2^1$.
You can also observe that 

$$\sqrt{2\sqrt{2\sqrt{2 ...\sqrt{2}}}} \cdot \sqrt{\sqrt{\sqrt{...\sqrt{2}}}}=2$$

thus

$$\sqrt{2\sqrt{2\sqrt{2 ...\sqrt{2}}}} =\frac{2}{\sqrt[2^n]{2}}$$