Prove that the splitting field of $x^{p}-q$ for prime numbers $p,q$ is an extension of degree $p(p-1)$ in $\mathbb{Q}$.
I know that the degree of the splitting field is bounded by $p!$, but I don't know how to prove that it equals $p(p-1)$ exactly. Any help would be appreciated.
Let $\zeta = e^{2\pi i/p}$ and $\omega = q^{1/p}$. Note that for $0 \leq j \leq p-1$, $\zeta^{j}\omega$ is a root of $x^p-q$. Since these are $p$ distinct roots and the polynomial is monic of degree $p$, we must have \begin{gather*} x^p-q=\prod_{j=0}^{p-1}(x-\zeta^{j}\omega). \end{gather*} From this, it is pretty easy to see that $\mathbb{Q}(\omega, \zeta)$ is a splitting field of the polynomial. You can show $[\mathbb{Q}(\omega):\mathbb{Q}]=p$ with Eisenstein's criterion and then note that $\zeta$ is a root of $x^{p-1}+ \cdots + x+1$ which is irreducible by the standard trick of applying Eisenstein's criterion to $(x+1)^{p-1}+\cdots + (x+1) + 1$, so that $[\mathbb{Q}(\zeta):\mathbb{Q}]=p-1$. Since gcd$(p,p-1)=1$, the degrees of the extensions multiply.