$k$ is an algebraic number field, and $\cal O$ is the ring of integers, $\cal O^\times$ is the set of invertible elements of $\cal O$.
Suppose $u\in\cal O^\times$ is not a $n$-th power. How can I choose $\frak p\unlhd\cal O$ prime so $u$ becomes a $n$-th power (mod $\frak p$)?
For $u=1$, the answer is a generalization of Dirichlet's Prime Number Theorem: if Fr$_{\frak p}$ fixes $k(\zeta_n)$ then $\zeta_n^{|\cal O/\frak p|}=\zeta_n$, so $\zeta_n^{|(\cal O/\frak p)^\times|} = 1$. In particular, $n\mid|(\cal O/\frak p)^\times|$, and so 1 is an $n$-th power in $\cal O/\frak p$.
How to generalize this to arbitrary $u\in\cal O^\times$?
Converse result: when $u = 1$, Lemma 3 in "Le probleme des groupes de congruence pour SL_2" by Serre states that, when $\zeta_n\notin k$, then we can find a prime so 1 is NOT an $n$-th power (mod $\frak p$):
1) Take $\sigma\in\mathrm{Gal}(k(\zeta_n)/k)$ so it acts on $\zeta_n$ non-trivially.
2) Using e.g. Chebotarev's theorem, find a prime $\frak p$ whose Frobenius Fr$_{\frak p}$ acts on $k(\zeta_n)$ same way as $\sigma$.
3) Since Fr$_{\frak p}$ fixes $\cal O/\frak p$ but moves $\zeta_n$, then $\sqrt[n]{1}=\zeta_n\notin\cal O/\frak p$.
This converse would transfer almost verbatim to $\sqrt[n]{u}$ for other $u\in\cal O^\times$; however, the question I asked does not.