 Solution

Suppose first that $L=K(\alpha )$ and the degree-$n$ minimal polynomial of $\alpha$ is an Eisenstein polynomial. It is irreducible, $\alpha$ is a uniformizer of $L$, and its valuation shows that $e(L/K)\ge n=[L:K]$. Equality follows, so $L/K$ is a totally ramified extension.

Conversely, suppose $L/K$ is totally ramified of degree $n$ and choose a uniformizer $\alpha$ of $L$. The field $K(\alpha )$ already has ramification index at least $n$, so it equals $L$. Letbe the minimal polynomial. All conjugates of $\alpha$ have $L$-valuation one. Each $a_i$ with $i<n$ is an elementary symmetric polynomial in products of at least one conjugate and therefore has positive $L$-valuation. Since valuations of elements of $K$ are multiples of $n$, every $a_i$ lies in the maximal ideal of ${\mathcal{O}}_K$. Moreoverso $a_0$ is not divisible by the square of that ideal. Thus $f$ is Eisenstein, proving the Eisenstein generator of a totally ramified extension criterion.