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.

Let us denote your sequence by $x_n$. Then
$$
x_1=\sqrt{2},\ x_{n+1}=\sqrt{2x_n} \quad \forall\ n \ge 1.
$$
Clearly $\sqrt{2}\le x_n<2$ for every $n \ge 1$, and 
$$
x_n-x_{n+1}=x_n-\sqrt{2x_n}=\frac{x_n^2-2x_n}{x_n+x_{n+1}}=\frac{x_n(x_n-2)}{x_n+x_{n+1}}<0 \quad  \forall n \ge 1.
$$
Hence $(x_n)$ is increasing and bounded above. It follows that $(x_n)$ is convergent. If $l$ denotes its limit, then $\sqrt{2}\le l \le 2$ and $l=\sqrt{2l}$. Solving the equation $l=\sqrt{2l}$ we get $l\in \{0,2\}$, and since $\sqrt{2} \le l \le 2$, we deduce that $l=2$.
