A number-field unit is exactly an algebraic integer of norm plus or minus one

Fix a $\mathbb{Z}$-basis of ${\mathcal{O}}_K$ and let $M$ be the matrix of the $\mathbb{Q}$-linear map $m_u:K\to K$, $m_u(x)=ux$, in that basis (Coordinate columns $[v{]}_{\mathcal{B}}$ and matrices $[T{]}_{\mathcal{B}}^{\mathcal{C}}$ of linear maps relative to ordered bases). Since $u\in {\mathcal{O}}_K$ and ${\mathcal{O}}_K$ is closed under multiplication, $u{\mathcal{O}}_K\subseteq {\mathcal{O}}_K$; hence every column of $M$ is the coordinate column of an element of ${\mathcal{O}}_K$, so $M$ has entries in $\mathbb{Z}$, and $N_{K\mathrm{/}\mathbb{Q}}(u)=\det M$.

Suppose first that $u$ is a unit of ${\mathcal{O}}_K$, so $u^{-1}\in {\mathcal{O}}_K$. The inverse of $m_u$ is $m_{u^{-1}}$, and its matrix in the same basis is $M^{-1}$; by the argument of step 1.1 with $u$ replaced by $u^{-1}$ this matrix has integer entries. Thus $M$ is invertible over $\mathbb{Z}$, so by [F2] $\det M$ is a unit of $\mathbb{Z}$, and [F4] gives $\det M=\pm 1$, that is, $N_{K\mathrm{/}\mathbb{Q}}(u)=\pm 1$.

Suppose conversely that $N_{K\mathrm{/}\mathbb{Q}}(u)=\det M=\pm 1$. Then $M$ is invertible and $M^{-1}=\det (M{)}^{-1}\mathrm{adj}(M)$ by [F3]; since $\det M=\pm 1$ is a unit of $\mathbb{Z}$ and the adjugate of an integer matrix has integer entries, $M^{-1}$ has integer entries. For every $x\in {\mathcal{O}}_K$ the coordinate column of $u^{-1}x$ is $M^{-1}$ applied to the coordinate column of $x$, hence is integral, so $u^{-1}{\mathcal{O}}_K\subseteq {\mathcal{O}}_K$; taking $x=1$ and using $1\in {\mathcal{O}}_K$ gives $u^{-1}\in {\mathcal{O}}_K$. Therefore $u\cdot u^{-1}=1$ exhibits $u$ as a unit of ${\mathcal{O}}_K$ together with its inverse $u^{-1}$.