This was stated in an algebraic number theory text
Let $K/ \mathbb{Q}$ be an algebraic number field (finite separable extension). Let $O$ be the integral closure of $\mathbb{Z}$ in $K$. Let $a \in O$, and deifne $L_a: K \rightarrow K$ by $x \mapsto xa$. Then $|O:aO| = |\det (L_a)|$.
EDIT: As from comments.
$O$ is a free $\mathbb{Z}$ module. $aO$ is also a free $\mathbb{Z}$. By Smith Normal Form $\{w_i \}$ be basis of $O$, and $\{a_i w_i \}$ be basis of $aO$. Then exists matrix $A \in GL_n(\mathbb{Z})$, such that $a \cdot $w $= A$ aw, where a is diagonal matrix with entires $\{a_i\}$. Thus $|\det (L_a)|= |\det (A)||\det \mathbf{a}| = \prod |a_i|$ which is what we wanted.