Let $D$ be a Euclidean domain, and let $\mathfrak{a} \subseteq D$ be a nonzero ideal. We show that $\mathfrak{a}$ is principal. Let$$ A = \{\nu(x) : x \in \mathfrak{a}, x \neq 0\}$$ be the set of Euclidean valuations of the non-zero elements of $\mathfrak{a}$ . Since $A$ is a non-empty set of non-negative integers, it has a minimum $m$ . Choose $d\in \mathfrak{a}$ such that $\nu(d) = m$ . Claim that $\mathfrak{a} = (d)$ . Clearly $(d) \subseteq \mathfrak{a}$ . To see the reverse inclusion, choose $x\in \mathfrak{a}$ . Since $D$ is a Euclidean domain, there exist elements $y,r\in D$ such that$$ x = yd + r$$ with $\nu(r) < \nu(d)$ or $r = 0$ . Since $r \in \mathfrak{a}$ and $\nu(d)$ is minimal in $A$ , we must have $r = 0$ . Thus $d \lvert x$ and $x\in(d)$ .