Let $A$ be a noetherian ring, and $\mathfrak{a}\subseteq A$ an ideal. Then there exists an $n\in \mathbb{N}$ such that
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$r(a{)}^n\subseteq \mathfrak{a}$
In particular, the nil-radical is nilpotent.

PROOF:

• Use that $r(\mathfrak{a})$ is finitely generated since $A$ is noetherian.