= Solution
A <composition series> is a chain $0=V_0\subsetneq V_1\subsetneq\cdots\subsetneq V_r=V$ of submodules with every factor $V_i/V_{i-1}$ simple; its length is $r$. Suppose $0\subset S\subset V$ is one of length two, and put $T=V/S$.
For any nonzero proper <submodule> $U$, either $U\cap S=S$ or $U\cap S=0$, by simplicity of $S$. In the first case, simplicity of $V/S$ forces $U=S$. In the second, the quotient map embeds $U$ nontrivially into $T$, so $U\cong T$ and its <image> is all of $T$. Hence $V=S\oplus U$, and $V/U\cong S$. Thus every nonzero proper <submodule> is simple and has simple quotient; every <composition series> has length two.
The <Jordan–Hölder theorem> states that any two <composition series> of a finite-length module have equal length and the same multiset of simple factors, up to isomorphism and order. The preceding argument proves this for length two: either $U=S$, giving factors $S,T$, or $U\ne S$, giving factors $T,S$.
The map $A\to S$, $a\mapsto ax$, is a surjective <module homomorphism> with <kernel> $I=\operatorname{Ann}_A(x)$. The <first isomorphism theorem for modules> gives $A/I\cong S$ as left modules; $I$ need only be a <left ideal>. Every <simple module> is cyclic, since any nonzero $x$ generates a nonzero <submodule>. As a quotient of the finite-dimensional regular module $A$, it is finite-dimensional. Extend $I$ to a <composition series> of $A$; its simple quotient $A/I$ is one of the factors. Jordan-Hölder shows that its isomorphism type belongs to the fixed finite list of factors of any one <composition series> of $A$. Hence \b[there are finitely many simple-module isomorphism types].
For the final assertion let $W$ be the <image> of $\psi$. Its <kernel> is $\bigcap I_j$. We prove $W=\prod S_j$ by induction on $k$. It is true for $k=1$. Projection onto the first $k-1$ factors is onto by induction. Its <kernel> on $W$ is a <submodule> of $S_k$, so is zero or $S_k$. If zero, $W\cong\prod_{j<k}S_j$. But its projection onto $S_k$ is onto, making $S_k$ a <composition factor> of that product, contrary to Jordan-Hölder and the pairwise distinct types. Thus the <kernel> is all of $S_k$. Surjectivity on the first factors together with this <kernel> proves $W$ is the entire product. The first isomorphism theorem now gives
$$
\boxed{A/\bigcap_{j=1}^kI_j\cong S_1\times\cdots\times S_k}
$$
as left $A$-modules, without asserting that the left ideals are two-sided.
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