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

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?

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 $x =2$.

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 $x=2$.