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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 $a_{n+1}=\sqrt{2}^{a_n}\to ... 2 Define$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 ... 18 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 ...