Let $u\in \mathbb{Z}_p^*$ be a unit in the ring of $p$-adic integers. Assume that $u^{1/p}\not\in \mathbb{Q}_p$, in other words $u$ is not a $p$-power.
I am wondering how the polynomial $f=x^p-u$ factors over $\mathbb{Q}_p$. I managed to prove (without using that $u$ is a unit) that f is irreducible if and only if it has no roots.
However, if $f$ has a root $\alpha$, then we have $$ f(x)=(x-\alpha)g(x) $$
I am wondering, do we know if $g$ is irreducible over $\mathbb{Q}_p$ or how it factors?
I claim that $f$ is irreducible, independent of concrete choice of $u$.
Note that we know all the roots of $f$: They are of the form $\zeta_p^k u^{1 / p}$ for $\zeta_p$ a primitive $p$th root of unity, $u^{1 / p}$ an initial choice of root and $k = 1, \ldots, p$. Let now $K = \mathbb{Q}_p(\zeta_p)$. Clearly $L = K(u^{1 / p})$ is a splitting field for $f$, and $\operatorname{Gal}(L / K) \cong \mathbb{Z} / p \mathbb{Z}$ generated by $u^{1 / p} \mapsto \zeta_p u^{1 / p}$ (to make this completely rigorous you could apply to Kummer theory) which implies that $f$ is irreducible over $K$ and hence also over the smaller field $\mathbb{Q}_p$ (since $f$ clearly has no roots in $K$ and $\operatorname{Gal}(L / K)$ acts transitively on its roots).
Note also that this does not use that $u$ is a $p$-adic unit, or in fact anything about the ground field except that it is not of characteristic $p$ (this assumption is hidden away in my appeal to Kummer theory), so the proof translates freely to other settings.