I'm going over some notes that claim that any extension of $\mathbb{Q}_p$ of degree $n$ has the form $\mathbb{Q}_p[\sqrt[n]{a}]$ for some $a\in\mathbb{Z}_p$. It references Proposition III.12 from Serre's "Local Fields", which states, more or less, that if $F$ is an extension of $\mathbb{Q}_p$ of degree $n$, then there is a $\mathbb{Q}_p$-basis of $F$ given by $1,x,x^2,\dots,x^{n-1}$ for some $x$.
What I don't understand is how to show that this can be used to produce an $n$th root. All we know is that there is a monic polynomial of degree $n$ such that $f(x)=0$. Does that really give us some $n$th root of an element of $\mathbb{Z}_p$?
If you track the reference to Serre's book, you'll see the actual result cited from there is that for every finite extension $K/\mathbf Q_p$ the ring of integers of $K$ has a power basis: $\mathcal O_K = \mathbf Z_p[\alpha]$ for some $\alpha$. That does not mean $K = \mathbf Q_p(\sqrt[n]{a})$ and it is stronger than your "more or less" comment that $K$ has a power basis over $\mathbf Q_p$! There was some kind of confusion between a $\mathbf Z_p$-basis of the form $\{1,\alpha,\ldots,\alpha^n\}$ and a $\mathbf Q_p$-basis of $K$ consisting of powers of the specific kind of number $\sqrt[n]{a}$.
That the ring of integers of every $p$-adic field has a power basis over $\mathbf Z_p$ is a bit surprising from the viewpoint of number fields, where the ring of integers always has a $\mathbf Z$-basis but need not have a power basis over $\mathbf Z$, a standard example being the cubic field $\mathbf Q(\gamma)$ where $\gamma$ is a root of $x^3-x^2-2x-8$.