= Solution
\b[The <valuation ring> and local compactness.] The opening, unlettered requests use
$$
\boxed{\mathcal O_K=\{x:|x|\leq1\},\quad
\mathfrak m=\{x:|x|<1\},\quad k=\mathcal O_K/\mathfrak m.}
$$
The 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 $|x/\pi|\leq1$, so
$$
\boxed{\mathfrak m=\pi\mathcal O_K,\qquad |k|=q<\infty.}
$$
Successive division by $\pi$ shows that every nonzero element of $\mathcal O_K$ is a unit times a power of $\pi$: division must terminate because $|\pi|^j\to0$. 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>.
\b[The iterated-power limit.] Normalize the <discrete valuation> by $v(\pi)=1$, and write $q=\ell^f$, where $\ell$ is the <residue characteristic>. For $u\in1+\pi^s\mathcal O_K$ with $s\geq1$, the binomial theorem gives
$$
v(u^q-1)\geq s+1.
$$
The 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 $2s\geq s+1$.
For $a\in\mathcal O_K^\times$, reduction gives $a^{q-1}\in1+\mathfrak m$, because $k^\times$ has order $q-1$. Iterating the inequality yields
$$
v\bigl(a^{(q-1)q^j}-1\bigr)\geq j+1,\qquad
v\bigl(a^{q^{j+1}}-a^{q^j}\bigr)\geq j+1.
$$
The 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 $\omega(a)^q=\omega(a)$, whence
$$
\boxed{a^{q^j}\longrightarrow\omega(a),\qquad
\omega(a)^{q-1}=1,\qquad\omega(a)\equiv a\pmod\pi.}
$$
This 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
$$
\mathcal O_K^\times=\mu_{q-1}\times(1+\mathfrak m).
$$
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