Prove that a vertex with minimum degree necessarily a vertex cut

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From Graphs and Digraphs (7 ed.), proof of Theorem 4.6:

If $G$ is not complete, then take a vertex $v$ with $\deg{v} = \delta(G)$ and note that $N(v)$ is a vertex-cut of $G$.

This is stated without proof, and I haven't been able to find any theorems about this in the book nor online. Why is it that a the neighborhood of a vertex of minimum degree is necessarily a vertex cut of its parent graph?