[proofplan]
We construct an explicit topological field isomorphism $k_K((T)) \to K$. Since $K$ has equal characteristic $p > 0$ and $k_K$ is perfect, the Teichmüller lift $[-]: k_K \to \mathcal{O}_K$ is a ring homomorphism (in characteristic $p$, the Teichmüller map is the unique multiplicative section that is also additive). We fix a uniformizer $\pi$ and show the map $\sum a_n T^n \mapsto \sum [a_n] \pi^n$ is a well-defined ring isomorphism $k_K[[T]] \to \mathcal{O}_K$, then extend to fraction fields.
[/proofplan]
[step:Recall the Teichmüller map in equal characteristic]
Since $\operatorname{char}(K) = p$ and $k_K$ is perfect, the Teichmüller map
\begin{align*}
[-]: k_K &\to \mathcal{O}_K
\end{align*}
is a ring homomorphism: it satisfies $[ab] = [a][b]$ (multiplicativity, which holds in all characteristics) and $[a + b] = [a] + [b]$ (additivity, which holds because $\operatorname{char}(\mathcal{O}_K) = \operatorname{char}(K) = p$ and the Teichmüller lift in characteristic $p$ is the unique section of the reduction map $\mathcal{O}_K \to k_K$ that is a ring homomorphism). Moreover, $[a] \equiv a \pmod{\mathfrak{m}_K}$ and $[0] = 0$, $[1] = 1$.
[guided]
Why is the Teichmüller map additive in equal characteristic $p$? In mixed characteristic, the Teichmüller representatives are only multiplicative. But when $\operatorname{char}(K) = p$, the residue field $k_K$ embeds into $\mathcal{O}_K/\mathfrak{m}_K$ as a subfield, and since $k_K$ is perfect, the Frobenius $x \mapsto x^p$ is an automorphism of $k_K$. The Teichmüller lift $[a] = \lim_{n \to \infty} \hat{a}^{p^n}$ (where $\hat{a}$ is any lift of $a$) stabilises and equals the unique $(p^n)$-th root of $\hat{a}^{p^n}$ in $\mathcal{O}_K$. In characteristic $p$, the Frobenius is a ring homomorphism, so the limit is additive: $[a + b] = [a] + [b]$. This gives $[-]$ the structure of a ring homomorphism (a section of $\mathcal{O}_K \twoheadrightarrow k_K$), making $k_K$ a coefficient field inside $\mathcal{O}_K$.
[/guided]
[/step]
[step:Construct the isomorphism $k_K[[T]] \to \mathcal{O}_K$]
Fix a uniformizer $\pi \in \mathcal{O}_K$ (i.e., $v(\pi) = 1$). Define the map
\begin{align*}
\Phi: k_K[[T]] &\to \mathcal{O}_K \\
\sum_{n=0}^{\infty} a_n T^n &\mapsto \sum_{n=0}^{\infty} [a_n] \pi^n.
\end{align*}
The series $\sum_{n=0}^{\infty} [a_n] \pi^n$ converges in $\mathcal{O}_K$ because $v([a_n]\pi^n) = v([a_n]) + n \geq n \to \infty$ (since $v([a_n]) \geq 0$ for all $a_n \in k_K$, with $v([a_n]) = 0$ when $a_n \neq 0$ and $v([0]) = \infty$), and $K$ is complete.
**$\Phi$ is a ring homomorphism:** Since $[-]$ is a ring homomorphism, $\Phi$ is the evaluation map $T \mapsto \pi$ on the ring $k_K[[T]]$ where $k_K$ is embedded in $\mathcal{O}_K$ via $[-]$. The evaluation map on a power series ring is a ring homomorphism whenever the series converge, which is guaranteed by $v(\pi) = 1 > 0$.
[guided]
More explicitly, for $f = \sum a_n T^n$ and $g = \sum b_n T^n$ in $k_K[[T]]$, the product $fg = \sum c_n T^n$ has $c_n = \sum_{j=0}^n a_j b_{n-j}$. Under $\Phi$:
\begin{align*}
\Phi(fg) = \sum_{n=0}^{\infty} [c_n] \pi^n = \sum_{n=0}^{\infty} \left[\sum_{j=0}^n a_j b_{n-j}\right] \pi^n = \sum_{n=0}^{\infty} \left(\sum_{j=0}^n [a_j][b_{n-j}]\right) \pi^n,
\end{align*}
where the second equality uses the additivity and multiplicativity of $[-]$. This equals $\Phi(f) \cdot \Phi(g)$ by the Cauchy product formula for convergent series in a complete valued ring.
[/guided]
[/step]
[step:Prove bijectivity using the $\pi$-adic expansion]
**Surjectivity:** Every element $x \in \mathcal{O}_K$ has a $\pi$-adic expansion using Teichmüller representatives as coset representatives. Specifically, define $a_0 \in k_K$ by $[a_0] \equiv x \pmod{\mathfrak{m}_K}$. Then $x_1 := (x - [a_0])/\pi \in \mathcal{O}_K$ (since $v(x - [a_0]) \geq 1$). Inductively define $a_n \in k_K$ by $[a_n] \equiv x_n \pmod{\mathfrak{m}_K}$ and $x_{n+1} := (x_n - [a_n])/\pi$. Then $x = \sum_{n=0}^{\infty} [a_n] \pi^n$, so $x = \Phi(\sum a_n T^n)$.
**Injectivity:** If $\Phi(\sum a_n T^n) = 0$, then $\sum [a_n] \pi^n = 0$. Suppose for contradiction that not all $a_n$ are zero, and let $n_0$ be the smallest index with $a_{n_0} \neq 0$. Then
\begin{align*}
0 = \sum_{n \geq n_0} [a_n] \pi^n = \pi^{n_0} \left([a_{n_0}] + \sum_{n > n_0} [a_n] \pi^{n - n_0}\right).
\end{align*}
Since $v(\pi^{n_0}) = n_0$ and $v([a_{n_0}] + \sum_{n > n_0} [a_n]\pi^{n-n_0}) = 0$ (as $v([a_{n_0}]) = 0$ and $v(\sum_{n > n_0} [a_n]\pi^{n-n_0}) \geq 1$, so the ultrametric inequality gives valuation $0$), we have $v(0) = n_0 < \infty$, a contradiction. Hence all $a_n = 0$ and $\Phi$ is injective.
Therefore $\Phi: k_K[[T]] \xrightarrow{\sim} \mathcal{O}_K$ is a ring isomorphism.
[/step]
[step:Extend to fraction fields and verify the topological isomorphism]
The isomorphism $\Phi: k_K[[T]] \xrightarrow{\sim} \mathcal{O}_K$ sends $T \mapsto \pi$, so it maps the uniformizer of $k_K[[T]]$ to the uniformizer of $\mathcal{O}_K$. Extending to fraction fields by localising at the uniformizer:
\begin{align*}
k_K((T)) = k_K[[T]][T^{-1}] \xrightarrow{\sim} \mathcal{O}_K[\pi^{-1}] = K,
\end{align*}
sending $\sum_{n \geq n_0} a_n T^n \mapsto \sum_{n \geq n_0} [a_n] \pi^n$ for $n_0 \in \mathbb{Z}$.
This is a topological isomorphism because both sides carry the topology induced by their respective discrete valuations ($T$-adic on $k_K((T))$ and $\pi$-adic on $K$), and $\Phi$ preserves valuations: $v_T(\sum a_n T^n) = \min\{n : a_n \neq 0\} = v_K(\sum [a_n]\pi^n)$. A valuation-preserving isomorphism of valued fields is automatically a homeomorphism.
Hence $K \cong k_K((T))$ as topological fields.
[/step]
