Let $R$ be a Dedekind domain, and $P$ a non-zero proper prime ideal of $R$. It is easy to show that we have proper descending chain of ideals $$R \supset P \supset P^2 \supset P^3\cdots$$ Also $R/P$ is a field, and so $P^k/P^{k+1}$ becomes module over $R/P$ i.e. vector space over field $R/P$.
Claim: $P^k/P^{k+1}$ is one dimensional vector space over $R/P$.
Proof: Consider $\pi\in P^k\setminus P^{k+1}$. Then $R\pi\subseteq P^k$ but $R\pi\nsubseteq P^{k+1}$. Thus, $R\pi$ factors into prime ideals as $R\pi=P^kQ_1\cdots Q_r$ where $Q_i$ are prime ideals different from $P$. Then $$R\pi+P^{k+1}=P^kQ_1\cdots Q_r + P^{k+1}=P^k(Q_1\cdots Q_r+P)=P^kR=P^k$$ where we used in last-but-second equality that $Q_1\cdots Q_r+P$ contains $P$ properly (easy to show) and hence is equal to $R$.
Thus $(R\pi+P^{k+1})/P^{k+1}=P^k/P^{k+1}$. [Next?]
Q. From this last equation, how can we conclude that $P^k/P^{k+1}$ is of dimension $1$?
In the book Basic Algebra by Knapp, I confused in similar arguments at the point above, there was some typo; see here, page 442.