Prove or disprove non-constructively there exist irrationals $a, b, c$ such that $a^{b^c}$ is rational.

Question

Consider the interesting question:

Do there exist irrationals $a$ and $b$ such that $a^b$ is a rational? Alternatively, prove or disprove that there exist irrationals $a$ and $b$ such that $a^b$ is a rational.

There has been an interesting nonconstructive proof:

Let $x:=\sqrt{2}^\sqrt{2}$. $x$ is either a rational or an irrational. (Law of excluded middle is used here.) If assuming that $x$ is rational, than its existence gives a proof. If assuming that $x$ is irrational, then we put $a:=x:=\sqrt{2}^\sqrt{2}$ and $b:=\sqrt{2}$ and we have $a^b=(\sqrt{2}^\sqrt{2})^\sqrt{2}=\sqrt{2}^2=2$, a rational. So, no matter we assuming x is a rational or not, there is always a proof.

A more interesting question:

Do there exist irrationals $a$ and $b$ and $c$ such that $a^{b^c}$ is a rational? Alternatively, prove or disprove that there exist irrationals $a$ and $b$ and $c$ such that $a^{b^c}$ is a rational.

Can we obtain a non-constructive proof, in which law of excluded middle is used, again?

Note: We follow the Tower rule ( https://brilliant.org/wiki/exponential-functions-properties/#the-tower-rule ) to compute $a^{b^c}=a^{(b^c)}$

Answer 1

Take $a=c=\sqrt2, b={\sqrt2}^{\sqrt2}$.

$$a^{b^c}={\sqrt2}^{({\sqrt2}^{\sqrt2})^{\sqrt2}}={\sqrt2}^{(\sqrt2)^2}=2$$

Answer 2

I'm not exactly sure if this is what you want, but off the top of my mind, by using $a^{\log_a b} = a$, you could come up with many examples of a rational number for $a^{b^c}$ with $a$, $b$, and $c$ being irrational, such as

$$\large{(\sqrt 2)^{e^{\ln 2}}}$$

or more generally,

$$\large{(\sqrt[n] x)^{a^{\log_a n}}}$$

for irrational values of $\sqrt[n] x$, $a$, and $\log_a n$.