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.
 A: 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 $l=2$.
A: You can also observe that 
$$\sqrt{2\sqrt{2\sqrt{2 ...\sqrt{2}}}} \cdot \sqrt{\sqrt{\sqrt{...\sqrt{2}}}}=2$$
thus
$$\sqrt{2\sqrt{2\sqrt{2 ...\sqrt{2}}}} =\frac{2}{\sqrt[2^n]{2}}$$
A: Put 
$$a_n:=\underbrace{\sqrt{2\sqrt{2...\sqrt 2}}}_{n\text{ square roots}}\Longrightarrow a_n=\sqrt{2a_{n-1}}$$
Since you've already proved the sequence $\,\{a_n\}\,$ converges, assume its limit is $\,x\,$ , so
$$x=\lim_{n\to\infty}a_n=\lim\sqrt{2a_{n-1}}=\sqrt{2x}\Longrightarrow x^2=2x$$
and since it's trivial that $\,x\neq 0\,$ then you can cancel above and get what you want.
A: Another proof:
Notice that:
$a_1 = 2^{1/2},\ a_2 = 2^{3/4},\ a_3 = 2^{7/8}$, and so on, thus,
$$a_n = 2^{(2^n-1)/2^n} = 2^{1-1/2^n}$$
Taking limit as $n$ tends to infinity, we have that
$$\lim_{n \rightarrow \infty} a_n = 2^1 = 2$$
A: 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 $2^1$. 
A: Hint : let $y = \sqrt{2.{y}}$ , Now solve for y.
A: It converges on $4$ if you use a different converger.
Written as $x=2\sqrt{x}$, as $\ln{x} =\dfrac{1}{2}\ln{x}+\ln{2}$, it converges on $x=4$, and so the proof is incomplete.
