= Solution
The ring form of the <Artin–Wedderburn theorem> states that the following conditions on a unital <ring> $R$ are equivalent: $R$ is a <semisimple ring>; $R$ satisfies the descending chain condition on <left ideals> and $J(R)=0$; and
$$
\boxed{R\cong\prod_{i=1}^s M_{n_i}(D_i),}
$$
where $s$ is finite, each $n_i\geq1$, and each $D_i$ is a <division ring>. The factors are unique up to permutation and isomorphism. The zero <ring> may be included as the empty product. This theorem also implies the corresponding right-handed statements.
First prove the implication from the chain condition and zero <Jacobson radical>. Among finite intersections of maximal <left ideals>, choose a minimal one, say $L=L_1\cap\cdots\cap L_t$. Such a minimal intersection exists by the descending chain condition. Intersecting with any other maximal <left ideal> cannot make $L$ smaller, so $L$ is contained in every maximal <left ideal>. Thus $L\subseteq J(R)=0$. It follows that the map of left <modules>
$$
R\longrightarrow\bigoplus_{j=1}^t R/L_j
$$
is injective. Each quotient is a <simple module>.
For completeness, a <submodule> $N$ of a finite <direct sum> $V=\bigoplus_j T_j$ of <simple modules> is itself a finite <direct sum> of <simple modules> and is a <direct summand> of $V$. Choose a sum $W$ of coordinate summands, maximal among those with $N\cap W=0$. For any omitted coordinate summand $T_j$, simplicity gives $T_j\cap(N+W)=0$ or $T_j\subseteq N+W$. In the first case $N\cap(W\oplus T_j)=0$, contradicting maximality; hence every coordinate summand lies in $N+W$. Therefore $V=N\oplus W$. Projection onto $N$ shows $N$ is a finite sum of images of the $T_j$, each image zero or simple. Add these images successively, discarding any image already in the sum: simplicity makes each new intersection zero. This yields a direct-sum decomposition. Applying this argument to the displayed injection proves that the left regular <module> is a finite sum of <simple modules>.
Conversely, if the regular <module> is a <direct sum> of <simple modules>, that sum is finite because $1$ lies in a finite subsum and generates the whole regular <module>. Every <left ideal>, being a <submodule>, has the preceding decomposition and splitting property. Its chain lengths are bounded by the number of simple summands, so the descending chain condition holds. The <Jacobson radical> annihilates every simple summand of the regular <module>, hence annihilates $1$ and is zero. This identifies the first two conditions.
Write the finite decomposition using pairwise nonisomorphic simple left <modules>:
$$
{}_RR\cong\bigoplus_{i=1}^s S_i^{n_i}.
$$
Every <simple module> is a quotient of $R$, by sending $1$ to any nonzero vector, so this list includes every simple isomorphism class. The <Schur lemma> gives $\operatorname{Hom}_R(S_i,S_j)=0$ for $i\ne j$, and $E_i=\operatorname{End}_R(S_i)$ is a <division ring>: any nonzero endomorphism has zero kernel and full image by simplicity. Consequently
$$
\operatorname{End}_R({}_RR)\cong\prod_{i=1}^sM_{n_i}(E_i).
$$
Every endomorphism of the left regular <module> is right multiplication by its value on $1$. Composition reverses multiplication of those values, so the left side is $R^{\mathrm{op}}$. Taking the <opposite ring>, and using transpose to identify $M_n(E)^{\mathrm{op}}$ with $M_n(E^{\mathrm{op}})$, gives the theorem's decomposition with $D_i=E_i^{\mathrm{op}}$. This keeps the left/right conventions explicit.
Conversely, the standard column <module> $D^n$ over $M_n(D)$ is simple: a nonzero coordinate of a vector can be inverted to construct a <matrix> sending it to any prescribed vector. The left regular <module> decomposes as the sum of its $n$ column ideals, each isomorphic to $D^n$. A finite product of these matrix <rings> is therefore a <semisimple ring>. This proves all implications. For uniqueness, the isomorphism classes of simple <modules> and their multiplicities in the regular <module> determine the numbers $n_i$ and the division <rings> $E_i^{\mathrm{op}}$. The multiplicity can also be recovered as the <dimension> of $\operatorname{Hom}_R(S_i,R)$ as a right $E_i$-<module>, so it does not depend on the chosen decomposition.
Now let $A=\mathbb C[G]$ for the finite <group> $G$. It has <dimension> $|G|$, with the elements of $G$ as a <basis>, so its <left ideals> satisfy the descending chain condition by their dimensions. The permitted assumption $J(A)=0$ makes the <Artin–Wedderburn theorem> applicable. Its decomposition is a complex-algebra decomposition
$$
A\cong\prod_iM_{n_i}(D_i),
$$
with each $D_i$ finite-dimensional over the central copy of $\mathbb C$. Each such <division algebra> is just $\mathbb C$. Indeed, for $d\in D_i$, finite dimensionality gives a nonzero polynomial relation $p(d)=0$ with complex coefficients. Factor $p$ into linear factors over $\mathbb C$. Their product at $d$ is zero, so the absence of <zero divisors> forces $d-\lambda=0$ for some scalar root $\lambda$. Thus every element is scalar.
The pairwise nonisomorphic simple <modules> of $\prod_iM_{n_i}(\mathbb C)$ are the column <modules> $\mathbb C^{n_i}$, with all other factors acting as zero. Hence $\dim_{\mathbb C}S_i=n_i$. Taking complex <dimensions> gives
$$
\boxed{\sum_{i=1}^k(\dim_{\mathbb C}S_i)^2=\sum_{i=1}^kn_i^2=\dim_{\mathbb C}\mathbb C[G]=|G|.}
$$
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