This is what I did:
Because $\mbox{rank}(A)=1$, then from rank nullity theorem $$\dim\ker A + \mbox{rank} A = n \implies \dim \ker A = n-1$$
and $gm(0)=dimE(0)=dimKerA=n-1$ where $E(0)$ is the eigenspace of the eigenvalue $0$.
So the characteristic polynomial of $A$ is: $C_A(x)=x^{n-1}(x-\lambda)$ where $\lambda$ could be $0$.
It is known that the sum of all eigenvalues of A equals the trace of A (Shown this using the fact that all matrices under $\mathbb C$ are triangularizable), so from that we'll conclude that $\lambda=0$ and $C_A(x)=x^n$
From here the minimal polynomial of A is $M_A(x)=x^k$ where $k\leq n$
From Cayley–Hamilton $M_A(A)=0 \implies A^k=0$
I'm not sure how to progress from here, I don't really know what I can say about the index of a nilpotent matrix when I know its rank. And I don't think I can say anything about the minimal polynomial given that I know the geometric multiplicity of the only eigenvalue
Best Answer
As noted in @Lost in Space answer, we can write $A = zw^T$, as a result of $\operatorname{rank}(A) = 1$. Now $0 = \operatorname{Tr}(A) = \operatorname{Tr}(zw^T) = \operatorname{Tr}(w^Tz) = w^Tz$ implies $w^Tz = 0$.
Therefore, $A^2 = zw^Tzw^T = z0w^T = 0$.