If $A$ is a simple finite dimensional $\mathbb{C}$-algebra then $A\cong M_n(\mathbb{C})$ I'm trying to prove the problem 17 of chapter 13 of the book Algebra: a Graduate Course (by Martin Isaacs), which is:

Let $A$ be a simple finite dimensional $\mathbb{C}$-algebra. Show that $A\cong M_n(\mathbb{C})$ for
some integer $n\geq 1$.

In one direction I'm trying to use the fact that $M_n(\mathbb{C})$ is generated by the $E_{ij}$ to construct the bijection with the generators of $A$ but I don't know if it is a good option.

The field $\mathbb{C}$ is not important. The statement is true for any algebraically closed field $F$.
Take a simple right $A$-module $V$. This is a finite dimensional vector space over $F$, because it is a quotient of $A$ modulo some maximal right ideal. Therefore its endomorphism ring is a finite dimensional division algebra over $F$. Since $F$ is algebraically closed the dimension must be $1$, so the endomorphism ring is $F$.
Also, the annihilator of $V$ in $A$ must be $\{0\}$, because $A$ is simple, so $V$ is faithful. Thus Wedderburn-Artin allows us to conclude that $A$ is isomorphic to $M_n(F)$, where $n=\dim_FV$ and the isomorphism is easily checked to be an $F$-algebra isomorphism.
