Prove convergence of $x_i$, where $x_1 = 1$ and $x_{n+1}=(\sqrt{2})^{x_n}$

Question

I need to prove that the sequence $\{x_n\}_{n=1}^\infty $ converges, where $x_1=1$ and $$x_{n+1}=(\sqrt{2})^{x_n} $$ The only way I can see to progress is to use the monotone bounded sequence theorem. (Nothing else makes sense to me.) My intuition tells me the sequence should be bounded above by $2$, and that $x_{n+1}>x_n$, but I can't work out what to do to prove those.

Answer 1

If $x_n < 2$, then

$$x_{n + 1} = \sqrt{2}^{x_n} < \sqrt{2}^2 = 2.$$

Furthermore, the function $x \mapsto 2^{x/2}$ is a strictly increasing function (its derivative is $2^{x/2} \cdot \ln \sqrt 2 > 0$, for example), and so $x_{n + 1} > x_n$.