If $x$ is a positive rational but not an integer, is $x^x$ irrational?
You have $y=x^x=\left(\cfrac pq\right)^{\frac pq}=\left(\cfrac {p^p}{q^p}\right)^{\frac 1q}$ so that $$y^q=\frac {p^p}{q^p}$$
Hence the rational fraction $y$ is a $p^{th}$ power - say $y=z^p$
Then $z^q=\frac pq$, and $\frac pq$ is a $q^{th}$ power. Now $q$ is an integer which is a $q^{th}$ power of an integer. But for $q\gt 1$ we have $2^q\gt q$.