Let $F_{\mathfrak{p}}$ be a completion of a number field $F$ at a non zero prime ideal $\mathfrak{p}$ of integer ring $\mathcal{O}$ and $\hat{\mathcal{O}}$ be its integer ring and $\hat{\mathfrak{p}}$ be its unique maximal ideal.
It is known that the relation between localized ring $\mathcal{O}_{\mathfrak{p}}$ and $\hat{\mathcal{O}}$ : $\mathfrak{p}\mathcal{O}_{\mathfrak{p}}$ and $\hat{\mathfrak{p}}$ are principal and have same generator.
I heard that $\hat{\mathfrak{p}}$ of $\hat{\mathcal{O}}$ equals to $\mathfrak{p}\hat{\mathcal{O}}$ (extension of ideal $\mathfrak{p}$ by inclusion $\mathcal{O}\to \hat{\mathcal{O}}$),but I don't know how to prove that.
To expand on @user8268 's comment, it is a standard result in $p$-adic fields that any $a\in F_\frak{p}$ can be written $a=\varepsilon\pi^e$ for some $e\in\Bbb{Z}$ and $\varepsilon$, where $\pi,\varepsilon\in \hat{\mathcal{O}}$ and $\varepsilon$ is a unit, while $\pi$ is a uniformizer, meaning that $\pi\in \hat{\frak{p}}\setminus (\hat{\frak{p}})^2$. In other words, it has valuation $1$, usually written $\nu_\hat{\frak{p}}(\pi)=1$. It is easy to check, from the definition of the valuation, that any element of $\frak{p}\setminus \frak{p}^2$ maps to a uniformizer, and combined with the fact that $\hat{\frak{p}}$ consists exactly of the elements of valuation at least $1$, the result follows.
The results above all follow directly from the properties of the valuation, turning $\hat{\mathcal{O}}$ into a discrete valuation ring (DVR).