Examples of $r^{sqrt{2}}$ as an irrational number, for real numbers $0

After I've read this question , and I tried some failed calculations, I've typed in Wolfram Alpha online calculator the code (sqrt(2)/2)^sqrt(2) to get as output that also this number is transcendent, and thus irrational.

Before this calculation with Wolfram Alpha, as I am saying was a curiosity that I was asking myself if it is possible deduce or are known some cases for which one can state that $r^{\sqrt{2}}$ is irrational, when $$0 $$\sqrt{2}\log r=\log p-\log q.\tag{1}$$
taking logarithms. And additionally if we presume, by contradiction, that $\sqrt{2}$ is a rational number, I can write the condition $$\frac{P}{Q}\log r=\log p-\log q,\tag{2}$$
where $P$ and $Q$ are positive integers satisfying $\gcd(P,Q)=1$.

But $(1)$ neither $(2)$ don't say nothing to me.

Question. Imagine that a friend ask me for a reasoning to get examples of irrational numbers of the form $$r^{\sqrt{2}},$$ when the real number $r$ runs on the set $0 what are simple requirements/conditions that need to be met those real numbers $0Thanks in advance.

Answer

The Gelfond-Schneider-theorem mentioned by Jack allows a more general choice for $r$ :

For every algebraic irrational $r$ with $0

A particular cute choice is the golden-ratio-number $\phi=\frac{\sqrt{5}-1}{2}$