Hint: what can you say about the Jacobson radical $J(A)$ in this case?
I think I understand how to begin this problem. Because we are considering a finite dimensional algebra over a field, A is finitely generated as a $\mathbb C$ module and is therefore Artinian. This implies $J(A)$ is a nilpotent ideal and therefore each element in $J(A)$ is nilpotent. If $J(A)$ is nonzero then we are done so without loss of generality assume $J(A)$ is the zero ideal. So $A$ is Artinian and Jacobson semisimple, therefore A is semisimple. Now we apply Artin–Wedderburn theorem and conclude A is a direct product of matrix algebras over division rings. If for any matrix algebra $M_n(D)$ we have $n>1$, we will get a non zero nilpotent element, so once again without loss of generality we can assume A is a direct product of division rings: $A = D_1 \times D_2 \times \ldots \times D_k$.
So my question is this: I am uncertain why, but i think since the complex numbers are an algebraically closed field, each copy of Di in the direct product is isomorphic to C. Is this correct? And if so why? Furthermore, this would imply $A = \mathbb C \times \mathbb C \times\ldots \times \mathbb C$ ten times, which has no nonzero nilpotent elements right? So this makes me think I’m doing something wrong. Thanks!
Also, sorry for the lame notation I’m new to this and in a hurry to understand this stuff before my exam which is in a couple days!
You are right, the $10$ dimensional $\mathbb C$-algebra $\mathbb C^{10}$ has no nonzero nilpotent elements.
So, the problem as stated is not correct. Check your source. Is it possible the statement additionally assumes the algebra is not commutative?
The only thing I can see to say about it is that it has dimension $9$ or less. The only use I can speculate for the hint is that apparently it was to get you on the track of showing the radical must be nonzero, but of course it isn't always nonzero.