TITLE: How can I prove $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}=2$
QUESTION [18 upvotes]: 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?

REPLY [7 votes]: 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 $x=2$.