= Solution
The <prime radical of a noncommutative ring> is $N(R)=\bigcap_P P$, with $P$ ranging over all <prime ideals of a noncommutative ring>. Here primeness means that for two-sided <ideals> $A,B$, $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$. A <minimal prime over an ideal> $I$ is a <prime ideal> containing $I$ with no strictly smaller <prime ideal> containing $I$.
The <ascending chain condition> on <left ideals> also applies to two-sided <ideals>. We first prove by <Noetherian induction> that every proper $I$ contains a finite product of <prime ideals>, each containing $I$. If there were counterexamples, choose a maximal one. It is not prime, so there are strictly larger two-sided <ideals> $A,B$ with $AB\subseteq I$. Neither is a counterexample; products of <prime ideals> contained in $A$ and in $B$ then concatenate to a product contained in $I$, a contradiction.
Write $P_1\cdots P_t\subseteq I\subseteq P_j$. Every <prime ideal> $P$ over $I$ contains at least one $P_j$, by primeness. The inclusion-minimal members of the finite list $P_j$ are therefore precisely the <minimal primes over an ideal> $I$: any smaller prime over $I$ would contain a member of that same list. Consequently there are finitely many. Replacing each factor $P_j$ by a minimal prime contained in it gives
$$
Q_1Q_2\cdots Q_t\subseteq I
$$
with each $Q_j$ minimal over $I$. Repeated factors may be necessary; in $k[\varepsilon]/(\varepsilon^m)$ the sole minimal prime $(\varepsilon)$ requires its $m$th power to reach zero.
Apply this to $I=0$. Since $N(R)\subseteq Q_j$ for every $j$,
$$
\boxed{N(R)^t\subseteq Q_1\cdots Q_t=0}.
$$
Thus the <prime radical> is a <nilpotent ideal>, as in <finite minimal primes and nilpotent prime radical>.
Now assume $R$ is commutative. A <prime ideal> $P$ is an <associated prime of a module> $M$ when $P=\operatorname{Ann}_R(m)$ for some $0\ne m\in M$. Every nonzero <module> over $R$ has an associated prime: the <ascending chain condition> gives a maximal annihilator of a nonzero element, and <maximal annihilator of a module element is prime>.
Since the given $M$ is a <Noetherian module>, repeatedly choosing such a cyclic <submodule> in the next quotient yields a finite <prime filtration>
$$
0=M_0\subsetneq M_1\subsetneq\cdots\subsetneq M_s=M,
\qquad M_i/M_{i-1}\cong R/P_i.
$$
The process terminates by the <ascending chain condition>. In a <short exact sequence> $0\to U\to M\to W\to0$, one has $\operatorname{Ass}(M)\subseteq\operatorname{Ass}(U)\cup\operatorname{Ass}(W)$. Indeed, if $P=\operatorname{Ann}(m)$ and $Rm\cap U=0$, the image of $m$ in $W$ still has annihilator $P$. Otherwise choose $0\ne am\in U$; then $a\notin P$, and primeness gives $\operatorname{Ann}(am)=P$. As $\operatorname{Ass}(R/P_i)=\{P_i\}$, induction along the <prime filtration> proves
$$
\boxed{\operatorname{Ass}_R(M)\subseteq\{P_1,\ldots,P_s\}}.
$$
In particular there are only finitely many <associated primes of a module>.
Finally let $P$ be a <minimal prime ideal> of $R$. Its <localization at a prime ideal> $R_P$ has exactly one prime, its <maximal ideal> $PR_P$. The preceding nilpotence result makes this maximal ideal nilpotent. Choose $0\ne y\in R_P$ annihilated by $PR_P$, for example a nonzero element in its last nonzero power, or $1$ if $PR_P=0$. Since elements outside the maximal ideal are <units>, $\operatorname{Ann}_{R_P}(y)=PR_P$.
Write $y=r/s$. The element $r/1$ is nonzero and also has annihilator $PR_P$. Generate $P$ by $p_1,\ldots,p_\ell$. For each $i$, choose $u_i\notin P$ with $u_ip_ir=0$, and put $u=\prod_i u_i$. Then $m=ur$ remains nonzero after localization, while $Pm=0$. If $am=0$, localization forces $a/1\in PR_P$, and hence $a\in P$. Therefore $\operatorname{Ann}_R(m)=P$. This proves that <minimal primes are associated primes>.
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