 Minimal-right-ideal proof of radical nilpotence

In a right Artinian ring, the powers of its Jacobson radical stabilize at an ideal $I$ satisfying $I^2=I$. If $I\ne 0$, choose a minimal right ideal $K$ with $KI\ne 0$. Since $(KI)I=KI$, minimality gives $KI=K$. Choosing $k\in K$ with $kI\ne 0$ gives $K=kR$ by the same minimality. Consequently $k=kj$ for some $j\in J(R)$. The unit criterion for the Jacobson radical makes $1-j$ invertible, forcing $k=0$, a contradiction. This proof avoids incorrectly applying the finitely generated form of Nakayama lemma to an arbitrary Artinian module.