If $x$ is a positive rational number, but not an integer, then can $x^{x^{x^x}}$ be a rational number ?

We can prove that if $x$ is a positive rational number but not an integer, then $x^x$ can not be rational:

Denote $x=\dfrac{b}{a},(a,b)=1,x^x=\dfrac{d}{c},(c,d)=1,$ $$\left(\dfrac{b}{a}\right)^\dfrac{b}{a}=\dfrac{d}{c} \hspace{12pt}\Rightarrow \hspace{12pt}\left(\dfrac{b}{a}\right)^b=\left(\dfrac{d}{c}\right)^a \hspace{12pt}\Rightarrow \hspace{12pt}b^b c^a=d^a a^b$$ Since $(a,b)=1,(c,d)=1,$ we have $c^a\mid a^b$ and $a^b\mid c^a$, hence $a^b=c^a.$ Since $(a,b)=1$, $a^b$ must be an $ab$-th power of an integer, assume that $a^b=t^{ab},$ then $a=t^a,$ where $t$ is a positive integer, this is impossible if $t>1,$ so we get $t=1,a=1$, hence $x$ is an integer.

Then from Gelfond–Schneider theorem , we can prove that if $x$ is a positive rational number but not an integer, then $x^{x^x}$ can not be rational. In fact, it can not be an algebraic number, because both $x$ and $x^x$ are algebraic numbers and $x^x$ is not a rational number.

- Can we prove that $x^{x^{x^x}}$ is irrational?
- Can $x^{x^{\dots (n-th)^{\dots x}}}~(n>1)$ be rational?