= Solution
Let $R$ be a commutative <Artinian ring>. We prove finite <composition length> first, without assuming the conclusion that $R$ is <Noetherian>.
Every <prime ideal> is maximal. Indeed, an Artinian <integral domain> is a field: for $a\ne0$, stabilization of $(a^m)$ gives $a^m=a^{m+1}b$, and cancellation gives $ab=1$. There are only finitely many <maximal ideals>. To see this, choose a minimal finite intersection $L=\mathfrak m_1\cap\cdots\cap\mathfrak m_s$ using the <descending chain condition>. Minimality gives $L\subseteq\mathfrak m$ for any other maximal ideal $\mathfrak m$. Since $\mathfrak m_1\cdots\mathfrak m_s\subseteq L$, primeness forces one $\mathfrak m_i\subseteq\mathfrak m$, and maximality makes them equal. Thus the <Jacobson radical> $J$ is this finite intersection, and the <Chinese remainder theorem> gives
$$
R/J\cong\prod_{i=1}^sR/\mathfrak m_i.
$$
We also need nilpotence of $J$. Its powers stabilize; put $I=J^m$ at a stabilization stage, so $I^2=I$. If $I\ne0$, choose an ideal $L$ minimal subject to $IL\ne0$. Since $I(IL)=IL\ne0$, minimality gives $IL=L$. Choose $a\in L$ with $Ia\ne0$. Minimality applied to the subideal $Ra$ gives $L=Ra$. The equality $IL=L$ then yields $a=xa$ for some $x\in I\subseteq J$. But $1-x$ is a unit by the <unit criterion for the Jacobson radical>; hence $a=0$, a contradiction. Therefore $J^m=0$.
Each layer $J^i/J^{i+1}$ is an <Artinian module> over the finite product of fields $R/J$. Its components over those fields are finite-dimensional: an infinite-dimensional vector space has a strictly descending chain of subspaces. Each layer consequently has finite <composition length>. The finite filtration
$$
R\supset J\supset\cdots\supset J^m=0
$$
proves that $R$ itself has finite <composition length>, and therefore satisfies the <ascending chain condition> on ideals. \b[Every commutative Artinian ring is Noetherian.] The zero ring satisfies the conclusion trivially.
The relevant <Artin–Wedderburn theorem> says that a <semisimple ring> which is a <right Artinian ring> has the form
$$
\boxed{R\cong\prod_{i=1}^sM_{n_i}(D_i),}
$$
where the $D_i$ are <division rings> and the positive integers $n_i$ and division rings are unique up to permutation and isomorphism. The division rings need not be commutative or finite-dimensional over a field. Conversely every such finite product is semisimple and a <right Artinian ring>.
Here is a proof which also works if semisimplicity is initially expressed as $J(R)=0$. A nonzero right ideal contains a minimal nonzero <right ideal> $L$ by the <descending chain condition>; $L$ is a <simple module>. If $L^2=0$, then for every $l\in L$ and $r\in R$, $(lr)^2=0$, so $1-lr$ has inverse $1+lr$. The <unit criterion for the Jacobson radical> would give $L\subseteq J(R)=0$, a contradiction. Thus some $a\in L$ induces a nonzero map $L\to L$ by left multiplication. This map is an isomorphism, since $L$ is simple. Choose $e\in L$ with $ae=a$. Then $a(e^2-e)=0$ and injectivity on $L$ gives $e^2=e$. We obtain $L=eR$ and the splitting
$$
R_R=eR\oplus(1-e)R.
$$
For a right ideal $K$ containing $e$, the same argument splits $K=eR\oplus(K\cap(1-e)R)$. Repeatedly split a minimal right ideal out of the remaining summand. The remaining summands form a strictly descending chain of right ideals, so the process terminates. Thus
$$
R_R\cong\bigoplus_{i=1}^sS_i^{\oplus n_i}
$$
for pairwise nonisomorphic <simple modules> $S_i$.
A nonzero map between <simple modules> is an isomorphism: its kernel and image leave no alternatives. Thus $D_i=\operatorname{End}_R(S_i)$ is a <division ring>, and $\operatorname{Hom}_R(S_i,S_j)=0$ for $i\ne j$. This is the needed <Schur lemma>. Taking endomorphisms of the direct sum gives $\prod_iM_{n_i}(D_i)$. Finally, left multiplication identifies $R$ with $\operatorname{End}_R(R_R)$: a right-module endomorphism sends $r$ to $f(1)r$, and composition corresponds to multiplication in $R$, not its opposite. This proves the displayed ring isomorphism.
For the converse, the right regular module of $M_n(D)$ is the direct sum of its $n$ row modules. Each row module is simple, since elementary matrices and nonzero scalar inverses send any nonzero row to any prescribed row. A finite product therefore has a finite semisimple regular module and is a <right Artinian ring>. The simple-module multiplicities and their endomorphism division rings recover the factors, establishing uniqueness.
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