This Seems Like It Should Be True

A variation on a theme....

Either $x = \sqrt{2}^{\sqrt{2}}$ is rational or irrational. If it is rational then $x$ is our counterexample. If it is irrational, then $x^{\sqrt{2}} = \sqrt{2}^{2} = 2$ is our counterexample.

So while we've established that the given assertion is false we're not actually sure whether $x$ or $x^{\sqrt{2}}$ is a specific counterexample, (not that it matters as far as this question is concerned, but it would be interesting to know). As it turns out, by the Gelfond-Schneider theorem $x$ is transcendental, (and hence irrational).

Note that none of integers, rational or complex numbers are closed under exponentiation. Respective counterexamples are $2^{-1} = \dfrac{1}{2}, 2^{\frac{1}{2}} = \sqrt{2}$ and $\large i^{i} = e^{-\frac{\pi}{2}}.$ It is closed for $\mathbb{Z}^{+}$ though.

Technically, the number should be

${ \sqrt { 2 } }^{ { \sqrt { 2 } }^{ \sqrt { 2 } } }$

not

${ \left( { \sqrt { 2 } }^{ \sqrt { 2 } } \right) }^{ \sqrt { 2 } }$

I can't understand