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The intended statement uses a nontrivial Non-Archimedean absolute value. Under that convention choose $t$ with $0<|t|<1$. A compact neighborhood $C$ of zero contains an open ball of some radius $r>0$. For sufficiently large $N$, $t^N{\mathcal{O}}_K$ is contained in that ball. The valuation ring is closed, so $t^N{\mathcal{O}}_K$ is a closed subset of $C$ and is compact. Scaling shows that ${\mathcal{O}}_K$ itself is compact.

Its maximal ideal $\mathfrak{m}=\{x:|x|<1\}$ is open. The residue field $k={\mathcal{O}}_K/\mathfrak{m}$ is therefore discrete and compact, hence finite: the open cover by its singleton sets has a finite subcover. Since $\mathfrak{m}$ has finite index, its complement is a finite union of open cosets. Thus $\mathfrak{m}$ is also closed and compact.

The continuous absolute value on a field attains its maximum on $\mathfrak{m}$. Nontriviality supplies a nonzero element there, so this maximum is $\rho =|\pi |$ for some $\pi \ne 0$, with $0<\rho <1$. If $x\in \mathfrak{m}$, then $|x/\pi |\le 1$, so $\mathfrak{m}=\pi {\mathcal{O}}_K$.

For arbitrary $x\ne 0$, choose the integer $n$ for which ${\rho }^{n+1}<|x|\le {\rho }^n$. Then $\rho <|x/{\pi }^n|\le 1$. This element cannot lie in $\mathfrak{m}$, by maximality of $\rho$, so its absolute value on a field is one. Therefore $|x|={\rho }^n$. This proves discrete valuation from nontrivial local compactness:

The nontrivial valuation in the local compactness criterion is essential. If the trivial absolute value is admitted, the displayed source claim is false: take $\mathbb{Q}$ with that absolute value on a field. It is complete and discrete, hence locally compact, but its valuation ring and residue field are both $\mathbb{Q}$, which is infinite. The intended nontrivial case has been proved above.

The valuation ring and local compactness. The opening, unlettered requests useThe ultrametric inequality makes ${\mathcal{O}}_K$ a valuation ring; its elements outside $\mathfrak{m}$ are exactly the units, so $\mathfrak{m}$ is its unique maximal ideal, and $k$ is the residue field.

Here the customary nontrivial-valuation hypothesis for a local field is necessary. With the trivial absolute value, any infinite field is discrete and locally compact, but its residue field is infinite and there is no nonzero uniformizer. We use the nontrivial valuation in the conclusions below.

Local compactness first gives a compact neighborhood of zero. Choose a sufficiently small nonzero $t$ so that the closed set $t{\mathcal{O}}_K$ is contained in that compact neighborhood. Then $t{\mathcal{O}}_K$, and hence ${\mathcal{O}}_K$, is compact. Since $\mathfrak{m}$ is open, its cosets give a discrete compact quotient $k$, which must be finite. Write $|k|=q$. The subgroup $\mathfrak{m}$ now has finite index and is closed in ${\mathcal{O}}_K$, hence compact. The continuous function $|\cdot |$ attains on $\mathfrak{m}$ a maximum $c$ with $0<c<1$. Choose $\pi$ with $|\pi |=c$. Every $x\in \mathfrak{m}$ has $|x/\pi |\le 1$, soSuccessive division by $\pi$ shows that every nonzero element of ${\mathcal{O}}_K$ is a unit times a power of $\pi$: division must terminate because $|\pi {|}^j\to 0$. Thus the value group is discrete. A Cauchy sequence eventually lies in a translate of the compact ring ${\mathcal{O}}_K$, has a convergent subsequence, and consequently converges itself. These arguments prove the discrete valuation from nontrivial local compactness and the completeness of locally compact nontrivially valued fields.

The iterated-power limit. Normalize the discrete valuation by $v(\pi )=1$, and write $q={\ell }^f$, where $\ell$ is the residue characteristic. For $u\in 1+{\pi }^s{\mathcal{O}}_K$ with $s\ge 1$, the binomial theorem givesThe linear term gains a factor $q\in \mathfrak{m}$ (or is zero in positive characteristic), while all terms of degree at least two have valuation at least $2s\ge s+1$.

For $a\in {\mathcal{O}}_K^{\times }$, reduction gives $a^{q-1}\in 1+\mathfrak{m}$, because $k^{\times }$ has order $q-1$. Iterating the inequality yieldsThe sequence is Cauchy and converges to a unit $\omega (a)$. Since the reduction of $a^{q^j}$ is always $\bar{a}$, its limit has the same residue. Taking limits after shifting the sequence gives $\omega (a{)}^q=\omega (a)$, whenceThis is the Teichmuller representative. The Hensel lemma applied to $T^{q-1}-1$, whose derivative is a unit at every nonzero residue, shows uniqueness for each residue. Consequently the prime-to-$\ell$ roots give a canonical splitting