Let $p$ be a prime and $\mathbb{Z}_p=\{a_0+a_1p+a_2p^2+\cdots: 0\leq a_i<p-1\}$ ring of $p$ adic integers and $\mathbb{Q}_p$ be fraction field of $\mathbb{Z}_p$.
We have valuation on $\mathbb{Q}_p$ given by $v_p(x)=n\in \mathbb{Z}$ where $x$ is unique representation $x=a_np^n+a_{n+1}p^{n+1}+\cdots$.
We see that $\mathcal{O}=\{x: v_p(x)\geq 0\}=\mathbb{Z}_p$ is a valuation ring wih unique non zero prime ideal $\mathfrak{m}=\{x:v_p(x)\geq 1\}=p\mathbb{Z}_p$ with residue field $\mathcal{O}/\mathfrak{m}=\mathbb{Z}_p/p\mathbb{Z}_p\cong \mathbb{Z}/p\mathbb{Z}$.
Let $K$ be a finite extension of $\mathbb{Q}_p$. We can have any valuation on $K$, but to be compatible with field extension criterion, we consider the valuation given by $v_K(x)=\frac{1}{n}v_p(N_{K/\mathbb{Q}_p}(x))$ where $N_{K/\mathbb{Q}_p}(x)$ is product of conjugates of $x$ under embeddings of $K$ in seperable closure of $\mathbb{Q}_p$.
For this we have $\mathcal{O}_K=\{x: v_K(x)\geq 0\}$ and $\mathfrak{m}_K=\{x:V_K(x)\geq 1\}$.
Can we say that $\mathcal{O}_K/\mathfrak{m}_K$ is also a finite field just like $\mathcal{O}/\mathfrak{m}$ with out using anything else not mentioned here??