= Solution
Let $G=\operatorname{Gal}(L/K)$ and normalize $v_L$. In lower numbering,
$$
G_{-1}=G,
\qquad
G_s=\{\sigma\in G:v_L(\sigma(x)-x)\geq s+1\text{ for every }x\in\mathcal O_L\}
\quad(s\geq0).
$$
These are the <ramification groups>; $G_0$ is the <inertia group> and $G_1$ is the <wild inertia group>.
Because $\mathcal O_L=\mathcal O_K[\alpha]$, every $x\in\mathcal O_L$ is $P(\alpha)$ for some $P\in\mathcal O_K[X]$. The polynomial identity $P(Y)-P(X)=(Y-X)Q(X,Y)$ has integral coefficients. Consequently the inequality for $\alpha$ implies it for every $x$, and the converse follows by taking $x=\alpha$. Therefore
$$
G_s=\{\sigma\in G:v_L(\sigma(\alpha)-\alpha)\geq s+1\}.
$$
Since $L/K$ is a <Finite Galois extension>, the minimal polynomial factors as
$$
f(X)=\prod_{\sigma\in G}(X-\sigma(\alpha)).
$$
Differentiating and evaluating at $\alpha$ gives
$$
f'(\alpha)=\prod_{1\ne\sigma\in G}(\alpha-\sigma(\alpha)),
$$
and hence
$$
v_L(f'(\alpha))
=\sum_{1\ne\sigma\in G}v_L(\sigma(\alpha)-\alpha).
$$
For a fixed nonidentity $\sigma$, its valuation is exactly the number of integers $s\geq0$ for which $\sigma\in G_s$. Interchanging the two finite sums proves the <ramification-group sum for a monogenic integer ring>:
$$
v_L(f'(\alpha))=\sum_{s\geq0}(|G_s|-1).
$$
The extension is unramified exactly when $G_0=1$, which by this nonnegative sum is equivalent to $v_L(f'(\alpha))=0$, or $f'(\alpha)\in\mathcal O_L^\times$. Moreover $|G_0|=e(L/K)$, so the $s=0$ term is $e(L/K)-1$. Equality
$$
v_L(f'(\alpha))=e(L/K)-1
$$
holds exactly when $G_1=1$, which is exactly tame ramification.
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