I'm going straight for the question, so:
let $x^x = p \in \mathbb{P}$, then $x$ is irrational. The proof is obvious, $p \neq a^a$ for some integer $a$, so $x^x = (\frac{a}{b})^{\frac{a}{b}}$, suppose that $x$ is in the lowest form, so $a$ does NOT divide $b$, then $x^x = p$ implies that $(\frac{a}{b})^{\frac{a}{b}} = p$, so $(\frac{a}{b}) = \sqrt[a]{p^b}$, since $\sqrt[a]{p^b}$ is irrational we reached a contradiction. Good.
Let's now suppose that $x$ is algebraic, then:
$a_nx^n + a_{n-1}x^{n-1}+...a_1x + a_0 = 0$, where $a_k$ is an integer and $a_n$, $a_0$ is different from zero.
Now if we raise both sides by $x$ (can i do this)? we get:
$a_nx^{xn}+a_{n-1}x^{x(n-1)} +...a_1x^x + a_0^x = 0^x$
In conclusion: $a_np^n + a_{n-1}p^{n-1} +...+ a_1p + t = 0$ where $t$ is transcendental by Gelfond Schneider theorem. So we reached a contradiction since the sum of any integer with a trascendental cannot be zero.
BUT
Since $x$ is irrational, when i raised the polynomial to the power of $x$, the distribution is correct? A mean, i can do such a thing and raise every term to the power of $x$?
If this is incorrect, then at least i can say that:
Suppose $x$ is algebraic, by Gelfond-Schneider Theorem we conclude that $x^{x^x} = p^x$ is trascendental, which is more obvious but still cool, since trascendental numbers are cool. I did a similar question some time ago, but it's about the irrationality of $x$ and not the transcendence.