Trivially we know $2^2 = 4$, $3^3 = 27$ are integral solutions to the equation $x^x = n$ where $n$ is a positive integer. Most other solutions are likely to be irrational (eg $x^x = 5$).
But does there exist any positive integer $n$ that can be expressed as $x^x$ where $x$ is a rational number that is not an integer? Is it possible to prove that there can be no rational solutions?
Proof by reference to rational root theorem:
Let $x = p/q$ (with p and q coprime). Raise both sides of the equation to the power of $q/p$.
Then x is a solution to the equation $x = n$ ^ $(q/p)$ But the rational root theorem states that the only solutions to this equation are integers or irrational. This implies x is both rational (by definition) and not rational (by the theorem). Hence there can be no rational solution (proof by contradiction).