 Solution

A local field is nondiscrete and locally compact. In the non-Archimedean case its valuation ring is compact by the scaling argument in part (b). Any Cauchy sequence has a tail in a translate of a compact ball, so has a convergent subsequence; the Cauchy property forces the whole sequence to converge. Thus the field is complete. Part (b) gives a discrete valuation, a uniformizer $\pi$, and a finite residue field $k={\mathbb{F}}_q$, $q=p^f$.

We construct the coefficient field rather than presuming it exists. The polynomial $T^q-T$ has derivative $-1$ in characteristic $p$. Every residue element therefore has a unique polynomial root lift by Hensel's lemma. The set $S$ of these $q$ lifts is a field: qth powers preserve addition and multiplication, so sums and products remain polynomial root lifts; a nonzero lift has inverse $a^{q-2}$. Reduction identifies $S$ with ${\mathbb{F}}_q$. This is the finite coefficient field in positive-characteristic local fields.

For $x\in {\mathcal{O}}_K$, choose $a_0\in S$ with the same residue, write $x=a_0+\pi x_1$, and repeat with $x_1\in {\mathcal{O}}_K$. Completeness givesThe first nonzero coefficient determines the valuation, so the expansion is unique. Multiplying an arbitrary field element by a suitable power of $\pi$ reduces it to this case. Consequently substitution $T\mapsto \pi$ gives a valued-field isomorphismAddition and multiplication agree with those of Laurent series by convergence, and a nonzero series has a nonzero leading coefficient, establishing injectivity as well as surjectivity.

Conversely, ${\mathbb{F}}_q((T))$ is complete for its T-adic absolute value on a field. Its valuation ring ${\mathbb{F}}_q[[T]]$ is compact: fixing finitely many coefficients gives the finite partitions into balls, and the successive coefficient choices realize the inverse limit of the finite rings ${\mathbb{F}}_q[T]/(T^n)$. Equivalently it is the product of countably many finite discrete coefficient sets. Thus it is a non-Archimedean local field. Distinct $q$ give nonisomorphic valued fields because their residue fields have different sizes. Replacing $|T|$ by any number in $(0,1)$ gives an equivalent absolute value on a field.