Is $f = X^p - p^2$ irreducible over $\mathbb{Q}$ for $p \geq 3$ a prime number? Normally I would use Eisenstein, but that does not work in this case. I tried calculating $f(X+1)$ but that does not seem to give any progress. For the context, I am trying to determine the splitting field $\Omega_f$ over $\mathbb{Q}$ and the degree of this extension. I am guessing it should be $\Omega_f = \mathbb{Q}(\sqrt[p]{p^2},\zeta_p)$, with degree $p(p-1)$, but I'm stuck in the proof. Can someone confirm that this is the case?
In the case $f = X^p - p$, I have done the following. In $\mathbb{C}$ we see that the roots of $f$ are given by $\zeta_p^k \sqrt[p]{p}$ with $0 \leq k \leq p-1$. Now we look at the tower $$ \mathbb{Q} \subset \mathbb{Q}(\sqrt[p]{p}) \subset \mathbb{Q}(\sqrt[p]{p},\zeta_p) = \Omega_{f}. $$ The first has degree $p$ with minimum polynomial $X^p - p$ (irreducible, Eisenstein for $p$). The second has minimum polynomial $\Phi_p$ of grade $p-1$. We see that this is irreducible over $\mathbb{Q}(\sqrt[p]{p}) \subset \mathbb{R}$, since $\zeta_p \in \mathbb{C}$, and also all powers $\zeta_p^k \notin \mathbb{R}$ for $1 \leq k \leq p-1$. So the total grade is $p(p-1) = p^2 - p$. Is this correct?
A root of $X^p-p^2=0$ generates the field $\Bbb Q(\sqrt[p]p)$ which has degree $p$ over $\Bbb Q$ (using Eisenstein).