Proof: Choose any basis $\alpha_1,\ldots,\alpha_n$ of $K$ over $\Rats$ . 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 $\alpha_1,\ldots,\alpha_n$ with each $\alpha_i\in\mathcal{O}_K$ .
Consider the group homomorphism $$ \varphi:K\rightarrow \Rats^n:\gamma\mapsto(\operatorname{Tr}_\Rats^K(\gamma\alpha_1),\ldots,\operatorname{Tr}_\Rats^K(\gamma\alpha_n) $$ where $\operatorname{Tr}_\Rats^K$ is the trace from $K$ to $\Rats$ . Note that $\varphi$ is $1-1$ , since if $\gamma\neq 0$ and $\varphi(\gamma)=0$ , then $$ n=\operatorname{Tr}_\Rats^K(1)=\operatorname{Tr}_\Rats^K(\gamma\gamma^{-1})=\operatorname{Tr}_\Rats^K(\gamma\sum r_i\alpha_i)=\sum r_i \operatorname{Tr}_\Rats^K(\gamma\alpha_i)= $$ where the last equality holds since $\gamma\in\ker\varphi$ .

Hence $\varphi:\mathcal{O}_K\hookrightarrow \Ints^n$ , so $\mathcal{O}_K$ is finitely generated and torsion-free. It has rank $\geq n$ since the $\alpha_i$ are linearly independent, and rank $\leq n$ since it injects into $\Ints^n$ , so it has rank $n$ .