Let $R$ be a Dedekind domain, let $\mathfrak{p}$ be a nonzero prime ideal, and let $a\in\mathfrak{p}^r\setminus\mathfrak{p}^{r+1}$. Then $(a)\subseteq \mathfrak{p}^r$ and so $\mathfrak{p}^r\mid (a)$. Since this is in a Dedekind domain, this is equivalent to the existence of an ideal $\mathfrak{a}\vartriangleleft R$ such that $(a)=\mathfrak{p}^r\mathfrak{a}$.
I want to show that $\mathfrak{p}$ and $\mathfrak{a}$ are comaximal, i.e. $\mathfrak{a}+\mathfrak{p}=R$.
By multiplying by the ideal inverse, it suffices to show that $\mathfrak{p}^r\mathfrak{a}+\mathfrak{p}^{r+1}=\mathfrak{p}^r$, although I suspect there's a more direct way. The key idea should be that $\mathfrak{a}+\mathfrak{p}$ is the smallest ideal containing both $\mathfrak{a}$ and $\mathfrak{p}$, but I don't see how to use that $a\not\in\mathfrak{p}^{r+1}$. Can anyone spell this out?
Edit: I now want to show that $\mathfrak{a}\mathfrak{p}^r\cap\mathfrak{p}^{r+1}=\mathfrak{a}\mathfrak{p}^{r+1}$. Here's what I tried.
Given the result that Eric showed, we have $v_\mathfrak{p}(\mathfrak{a}+\mathfrak{p})=0$. Then $$v_\mathfrak{p}(\mathfrak{a}\mathfrak{p}^r\cap\mathfrak{p}^{r+1})=\max(v_{\mathfrak{p}}(\mathfrak{a}\mathfrak{p}^r), v_\mathfrak{p}(\mathfrak{p}^{r+1}))=v_\mathfrak{p}(\mathfrak{p}^{r+1})=r+1$$ Then since the intersection is the largest ideal contained in both, which must have $r+1$ prime factors, the intersection must be $\mathfrak{a}\mathfrak{p}^{r+1}$. Is there anything wrong here?
Since $\mathfrak{p}$ is a maximal ideal of $R$, $\mathfrak{a}+\mathfrak{p}$ can only be equal to either $\mathfrak{p}$ or $R$. If it is equal to $\mathfrak{p}$, then $\mathfrak{a}\subseteq\mathfrak{p}$, so $a\in\mathfrak{p}^r\mathfrak{a}\subseteq\mathfrak{p}^{r+1}$. This is a contradiction, so we must have $\mathfrak{a}+\mathfrak{p}=R$.