Let $K$ be a field and let $v:K\to G\sqcup\{+\infty\}$ be a valuation on $K$, where $G$ is a totally ordered abelian group. For each $\alpha$, define $V_\alpha=v^{-1}((\alpha,+\infty])$. Then $\{V_\alpha\}_{\alpha\in G}$ is an open neighborhood basis at zero of a unique topology $\mathscr{T}_v$ on $K$ (Bourbaki, Commutative Algebra, Ch. VI, § 5, no. 1). We can thus complete $K$ analytically (i.e., through the means of Cauchy filters; or Cauchy sequences if $\mathscr{T}_v$ is first-countable) and obtain a topological ring $\hat{K}$. Since $\mathscr{T}_v$ is well-behaved, $\hat{K}$ is actually a field (ibid., no. 3, Proposition 5 (a), or the link below for a proof when $\mathscr{T}_v$ is first-countable).
Recently a person told me that in the case $v$ is a discrete valuation one can construct $\hat{K}$ algebraically, via the following idea: if $\mathfrak{m}$ denotes the ideal of $v$, i.e., $\mathfrak{m}=v^{-1}((0,+\infty])$ (it is the maximal ideal of the local ring $A=v^{-1}([0,+\infty])$), then one consider the following inverse limit on topological abelian groups $$ \label{inv_lim}\tag{1} \hat{K}=\varprojlim_{n\geq 0} K/\mathfrak{m}^n $$ (where $\mathfrak{m}^0=A$). Note that this inverse limit a priori is not a ring, for $K/\mathfrak{m}^n$ are not rings, as $\mathfrak{m}^n$ is not an ideal of $K$. The same person told me that only a posteriori one can define ad-hoc a multiplication on $\hat{K}$, so that $\hat{K}$ becomes a topological field.
My only question is:
Where can I find the details of this construction? (Book, article, or link)
It is not that I don't believe in the completion of a field (I'm comfortable with the analytic completion via Cauchy sequences, and I understand why it's a topological field [ref]). However, I've been unable to find any mention on the literature of the inverse-limit approach to the completion of a field with respect to a discrete valuation, nor searching in Google or here in MSE.
