proof of the ring of integers of a number field is finitely generated over $\mathbb{Z}$

Proof: Choose any basis ${\alpha }_1,\mathrm{\dots },{\alpha }_n$ of $K$ over $\mathbb{Q}$. Using the theorem in the entry multiples of an algebraic number, we can multiply each element of the basis by an integer to get a new basis ${\alpha }_1,\mathrm{\dots },{\alpha }_n$ with each ${\alpha }_i\in {𝒪}_K$.
Consider the group homomorphism

$$\phi :K\to {\mathbb{Q}}^n:\gamma \mapsto ({\mathrm{Tr}}_{\mathbb{Q}}^K(\gamma {\alpha }_1),\mathrm{\dots },{\mathrm{Tr}}_{\mathbb{Q}}^K(\gamma {\alpha }_n))$$

where ${\mathrm{Tr}}_{\mathbb{Q}}^K$ is the trace (http://planetmath.org/trace2) from $K$ to $\mathbb{Q}$. Note that $\phi$ is $1-1$, since if $\gamma \ne 0$ and $\phi (\gamma )=0$, then

$$n={\mathrm{Tr}}_{\mathbb{Q}}^K(1)={\mathrm{Tr}}_{\mathbb{Q}}^K(\gamma {\gamma }^{-1})={\mathrm{Tr}}_{\mathbb{Q}}^K(\gamma \sum r_i{\alpha }_i)=\sum r_i{\mathrm{Tr}}_{\mathbb{Q}}^K(\gamma {\alpha }_i)=0$$

where the last equality holds since $\gamma \in \ker \phi$.

Hence $\phi :{𝒪}_K\hookrightarrow {\mathbb{Z}}^n$, so ${𝒪}_K$ is finitely generated and torsion-free. It has rank $\ge n$ since the ${\alpha }_i$ are linearly independent, and rank $\le n$ since it injects into ${\mathbb{Z}}^n$, so it has rank $n$.