Let $R$ be a commutative ring and $p \in R$. Consider the $p$-adic completion $\widehat{R} := \varprojlim_{n} \, R/p^n$.
When do we have $\widehat{R}/p^n \widehat{R} \cong R/p^n R$?
For fixed $n$ there are natural maps of $R$-algebras $R/p^n \to \widehat{R}/p^n \widehat{R} \to R/p^n$ whose composition is the identity of $R/p^n$. But I don't know why or if the other composition is the identity of $\widehat{R}/p^n \widehat{R}$.
If $R$ is noetherian, it seems to be true, by general results of completions: $\widehat{R}/p^n \widehat{R} = \widehat{R} \otimes_R R/p^n R = \widehat{R/p^n R} = R/p^n R$.
In general, it suffices to treat the case $n=1$. In fact, this implies $\widehat{R} = R + p \widehat{R}$ and then by induction $\widehat{R} = R + p^n \widehat{R}$ for all $n \geq 1$, i.e. $R/p^n \to \widehat{R}/p^n \widehat{R}$ is surjective.
For $n=1$ we have to show that $\widehat{R}/p \widehat{R} \to R/p$ is injective. But I haven't been able to prove this. At least when $p$ is regular there are no problems. Here is an elementary reformulation of the claim:
Given a sequence of elements $x_n \in p R$ such that $x_{n+1} \equiv x_n \bmod p^n$, why is there is a sequence of elements $y_n \in R$ such that $x_n \equiv p y_n \bmod p^n$ and $y_{n+1} \equiv y_n \bmod p^n$?
Actually, it's always the case that $\widehat{R}/p^n \widehat{R} = R/p^n R$.
To see this, recall that any element of $\widehat{R}$ can be written as a sum $\sum_{i = 0}^{\infty} p^i r_i,$ where $r_i \in R$ (or, more precisely, is in the image of $R$ in $\widehat{R}$). Thus we see that $\widehat{R} = \text{im}(R) + p^n \widehat{R},$ and so the natural map $R/p^n \to \widehat{R}/p^n \widehat{R}$ is surjective.
Combined with what you already know, this completes the proof.
[I think this extends to the completion w.r.t. any finitely generated ideal.]