I'm trying to prove the question above. I'm not sure whether this is true or not, but I'm trying to figure it out. If $A$ is semi-simple I don't think it's too hard to see that it's true, but otherwise, all I know is that $A$ has at most $n$ simple modules, and that every simple module is $1$ dimensional by Schur's lemma.
Not exactly sure how to proceed other than write out $$0= M_0 \subset M_1 \subset \dots \subset M_{n+1} = A $$ as a composition series, then, every simple module is isomorphic to $M_i/M_{i-1}$ and there are precisely $n+1$ of these for dimension reasons. Now if $M_i/M_{i-1} \equiv M_j/M_{j-1}$ we have to somehow prove that $i = j$.
On the other hand there could be a counterexample, but I can't come up with a finite dimensional $\mathbb{C}$ algebra that is both commutative and not semi-simple.
Would appreciate any help here.
This is false. For instance, if $A=\mathbb{C}[x]/(x^2)$, then $A$ is 2-dimensional but has only one simple module (since its only maximal ideal is $(x)$).
More generally, if $A$ is a finite-dimensional commutative $\mathbb{C}$-algebra and $N$ is its nilradical, then $A/N$ is semisimple and every maximal ideal of $A$ contains $N$. Thus since $A/N$ has $\dim A/N=\dim A-\dim N$ different simple modules, so does $A$. So the failure of $A$ to have $\dim A$ simiple modules is exactly measured by the nilpotent elements of $A$.