How to solve the following exercise:
Let $q$ be prime. Show that the set of primes p for which $p \equiv 1\pmod q$ and
$$2^{(p-1)/q} \equiv 1 \pmod p$$
has Dirichlet density $\dfrac{1}{q(q-1)}$.
I want to show that $X^q-2$ (mod $p$) has a solution and $q$ divides $p-1$ , these two conditions are simultaneonusly satisfied iff p splits completely in $\Bbb{Q}(\zeta_q,2^{\frac{1}{q}})$. $\zeta_q $ is primitive $q^{th}$ root of unity. If this is proved the I can conclude the result by Chebotarev density theorem.
For any finite extension $K\supset\mathbb Q$, a non-ramified prime ideal $\mathfrak P\subset O_K$ such that $O_K/\mathfrak P=\mathbb F_p$ is equivalent to an embedding $\phi:K\to\mathbb Q_p$ (when an embedding is given then $\mathfrak P=O_K\cap \phi^{-1}(p\mathbb Z_p)$; when $\mathfrak P$ is given then $K_{\mathfrak P}\cong\mathbb Q_p$ and the embedding is $K\subset K_{\mathfrak P}$).
[Other prime ideals containing $p$ (ramified and/or such that $O_K/\mathfrak P\cong\mathbb F_{p^f}$ with $f>1$) are equivalent to embeddings (with dense image) to finite extensions of $\mathbb Q_p$.]
Since your extension $K:=\Bbb{Q}(\zeta_q,2^{\frac{1}{q}})\supset\Bbb Q$ is Galois, a prime $p$ splits completely in $K$ iff there a unramified $\mathfrak P\subset O_K$ such that $O_K/\mathfrak P=\mathbb F_p$ (as the Galois group acts transitively on the prime ideals containing $p$), i.e. iff there is an embedding $K\to\mathbb{Q}_p$ , i.e. iff $\mathbb Q_p$ contains $q$-th roots of $1$ and a $q$-th root of $2$.
The condition $p \equiv 1\pmod q$ is equivalent to having $q$-th roots of $1$ in $\mathbb Q_p$, and solvability of $X^q=2$ in $\mathbb F_p$ is equivalent, for $p\neq 2,q$, to its solvability in $\mathbb Q_p$ (both by Hensel's lemma). The two conditions combined are thus equivalent to existence of an embedding $K\to\mathbb Q_p$, and we're done.