Q:

Proof that ideals of Dedekind domains are generated by at most two elements: how does the bolded step in the proof work?

Sorry for the slightly obscure title, but I could not think of a better way to word it. The lecture notes for my commutative algebra course have the following proof that Dedekind domains' ideals are always generated by at most two elements, but there is one step in the proof that neither I nor anyone else I've asked can make any sense of. I've asked people good enough at this that I'm convinced it's at least not blindingly obvious. 
Proof. Let $R$ be a Dedekind domain and let $I$ be an ideal of $R$. Let $a\in I\setminus\{0\}$, and let $J = Ra$. We can factorise $J$ as $\prod_{i=1}^m P_i^{s_i}$ for some prime ideals $P_i$ and $s_i\in \mathbb{N}$. Since $J\subseteq I$, we must have $I = \prod_{i=1}^mP_i^{n_i}$ for some integers $0\leq n_i \leq s_i$. 
We have $I/J \cong \prod_{i=1}^m P_i^{n_i}/P_i^{s_i}$ by the Chinese remainder theorem. Let us choose $b_i\in P_i\setminus P_i^2$. This gives $Rb_i^{n_i} + P_i^{s_i} = P_i^{n_i}$ and so each $P_i^{n_i}/P_i^{s_i}$ is a principal ideal in $R/P_I^{s_i}$. Hence $I/J$ is a principal ideal generated say by $b+J$ in the ring $R/J$. Then $I = Rb + J = \langle a, b \rangle$. $\quad\square$
Any help on parsing this proof would be greatly appreciated by me and several others. 

A:

$$P^s\subseteq b^nR+P^s\subseteq P^n$$ so the prime factorization of $b^nR+P^s$ is $P^k$ for some $k$. Now $$b^nR+P^s\not \subseteq P^{n+1}$$ as $$b^{n}\not \in P^{n+1}$$ so we can only have $k=n$.

