How can I prove $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}=2$ How can I prove 

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

I don't know which method can be used for this?
 A: Step One. Define the recursive sequence
$$
a_0=\sqrt{2}, \quad a_{n+1}=\sqrt{2}^{a_n},\,\,n\in\mathbb N.
$$
Step Two. Show that $\{a_n\}$ is increasing (inductively), and upper bounded by $2$ (also inductively).
Step Three. Due to Step Two the sequence $\{a_n\}$ is convergent. Let $a_n\to x$. Clearly, $\sqrt{2}<x\le 2$.
But $a_{n+1}=\sqrt{2}^{a_n}\to x$, as well. Hence 
$$
x=\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\sqrt{2}^{a_n}=\sqrt{2}^x.
$$
Thus $x$ satisfies $\sqrt{2}^x=x$.
Step Four. Show that $x<\sqrt{2}^x$, for all $x\in (\sqrt{2},2)$, and thus $x=2$.
A: We can define $x=\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}$ as follows:
Let $x_1 = \sqrt 2$ and $x_{n+1} = (\sqrt 2)^{x_{n}}$
We can show $x_n \lt 2\ \forall n$ by induction, since if $y \lt 2$, then $(\sqrt 2)^y \lt 2$. And $x_n$ is clearly monotonically increasing, so $x_n \to x$.
But $$x_{n+1} = (\sqrt 2)^{x_{n}}$$
so taking limits, we get that $$x = (\sqrt 2)^{x}$$
Solving this, and using the fact that $x \le 2$ gives $x =2$.
A: Define $x_1=\sqrt{2}$ and $x_{n+1}=\sqrt{2}^{x_n}$.
Prove by induction that $x_n \leq x_{n+1} \leq 2$. As the sequence is bounded and increasing, it is convergent, and the limit is between $x_1=\sqrt{2}$ and $2$. 
Finish the proof by observing that 
$$\sqrt{2}^x=x$$
has an unique solution on the interval $[\sqrt{2}, 2]$.
For the last part, as well as for the monotony, you should study first the monotony/sign of $x-\sqrt{2}^x$ on $  [\sqrt{2}, 2]$
