This question concerns Proposition on page 43 of Associative Algebras by R. S. Pierce.
Lemma c. Let $N$ be a right ideal of the algebra $A$ that satisfies $N^{k}=0$. If $P$ is a simple $A$-module, then $PN=0$. Moreover, $N\subseteq rad(A)$, where $A$ is considered as a right $A$-module.
Proposition. The following conditions are equivalent for minimal right ideals $N_{1}$ and $N_{2}$ of the semisimple algebra $A$. (i) $N_{1}\cong N_{2}$ as $A$-modules (ii) $N_{1}N_{2}\neq 0$. (iii) There is an element $x\in A$ such that $N_{1}=xN_{2}$.
Proof of Proposition (i) $\Rightarrow$ (ii): If $\phi:N_{1}\rightarrow N_{2}$ is an $A$-module isomorphism, then $\phi(N_{1}N_{2})=\phi(N_{1})N_{2}=N_{2}^{2}\neq 0$ by Lemma c.
My question is how does $N_{2}^{2}\neq 0$ follow from lemma c?
Any hints or answers is appreciated.
Since the algebra is semisimple, the Jacobson radical $rad(A)=\{0\}$. $N_2^2=\{0\}$ would imply $rad(A)\neq \{0\}$.