Isomorphism between rings $A/m^n \rightarrow \hat{A}/\hat{m}^n$. In Lemma 7.25, pg. 115, of Milnes' notes.

For $n \ge 1$, the map $A/m^n \rightarrow \hat{A}/\hat{m}^n$ is an isomorphism.

$A$ is ring associated to a discrete valuation $(K,|\quad|)$, $m$ its maximal ideal, $\hat{A}$ is that associated to its completion, $(\hat{K}, |\quad|)$.
I believe we have to prove that this is a homeomorphism. He claims that because $m^n$ is both open and closed we have bijectivity.
(i) How does this follow?
(ii) How does this show the topology are the same?
 A: I think you misread Milne. His claim is only that the map (let's call it $f$) is an isomorphism of rings, the only non-obvious part of which is bijectivity. Then, he claims that
(1) injectivity follows from $m^n$ being closed in $A$
(2) surjectivity follows from $\hat m^n$ being open in $\hat A$.
Ad (1): Notice that $m^n$ is dense in $\hat m^n$ and hence in $\ker(f) = \hat m^n \cap A$.
Ad (2): Let $\hat a \in \hat A$, and let $a_k$ be a sequence in $A$ with $\lim_{k\to \infty} a_k = \hat a$. Since $\hat m^n$ is open in $\hat A$, the set $\hat a +\hat m^n$ is an open neighbourhood of $\hat a$ in $\hat A$, hence all but finitely many $a_k$ are contained in it. For all these $a_k$, we have $f(a_k+m_n) = \hat a +\hat m_n$.
After that is set and done, you can of course also conclude that $f$ is a homeomorphism, which is kind of trivial once you've noticed that (because $m^n$ is open and closed in $A$, and $\hat m^n$ is open and closed in $\hat A$) the natural (quotient) topology on both sides is the discrete topology.
