Let $V = (V, +, \cdot, *)$ be an $n$-dimensional associative algebra over an algebraically closed field $F$. Let $\{v_1, \dots, v_n\}$ be a basis for $V$. I am asked to show that, for each $i = 1, \dots, n$, there are only two possible values that $v_i * v_i$ can take: either $0$ or $v_i$.
Actually the exercise I am trying to solve is only considering the case $n = 2$, but I guess the more general case is also true and not much more difficult.
Maybe related to this?
Since you posted the exact wording of the problem in the comments, we can establish that as it is written, the exercise is just false. It's pretty easy to see: take $V=F\times F$, and take the basis $v_1=(2,0)$ and $v_2=(0,2)$. Then clearly $v_i^2=2v_i$.
On the other hand, what is true, and probably what the exercise is about, is that there exists a basis such that $v_i^2=0$ or $v_i$. This is another way of saying that there are exactly two $2$-dimensional algebras over $F$, up to isomorphism: $F\times F$ and $F[\varepsilon]=F[x]/(x^2)$.
To see that, note first that there is a natural embedding $F\subset V$ since $V$ has a unit, and we can take $v_1=1$ as our first basis element. Then take any $x\in V\setminus F$, and check that it satisfies a quadratic equation with coefficient of $F$. Then try to see that up to an affine transformation, you can assume that this equation is either $x^2=0$ or $x^2=x$.