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The polynomial of $\alpha$ satisfies the Eisenstein criterion at $p$. Let $\mathfrak{p}$ be any prime ideal of ${\mathcal{O}}_K$ above $p$, normalize its ideal valuation by $v_{\mathfrak{p}}(\mathfrak{p})=1$, and write $e=v_{\mathfrak{p}}(p)$ for its ramification index of a prime ideal. Put $t=v_{\mathfrak{p}}(\alpha )$. Inthe constant term has valuation $e$, while every other $a_j{\alpha }^j$ has valuation at least $e+jt$. The non-Archimedean valuation property then forcesIndeed, either strict inequality would leave a unique term of least valuation in the equation. Since $t$ is a positive integer and $e\le n$, it follows that $t=1$ and $e=n$.

The fundamental inequality ${\sum }_{\mathfrak{p}\mid p}{e}_{\mathfrak{p}}{f}_{\mathfrak{p}}\le n$ now shows that $\mathfrak{p}$ is the unique prime above $p$ and has residue degree one. At this prime,while $(p,\alpha )$ has no factor away from $p$. Unique prime ideal factorization therefore givesFinally $v_P(\alpha )=1$, soThis is total ramification from an Eisenstein polynomial.