Is it possible to have a $2\times2$ matrix such that $A^3=0\ne A^2$? I know it is possible to have such a matrix if $A$ is $3\times 3$. I think it won't be possible in this case but I'm not sure how to prove a general case.
A specific case proof would work something like this.
$A$ is determined by its action on $e_1,e_2$
Say $Ae_1=\alpha e_1 + \beta e_2$.
Note that $\alpha=0$ else $A^3\ne0$
Similarly $Ae_2=\gamma e_1$
Hence both $\beta=0=\gamma$
How to prove for larger dimension? That is, $A\in M_{n\times n}$ then it is not possible that $A^n\ne0=A^{n+1}$

Consider the images of $A, A^2$ and $A^3$ as subspaces of the plane. If the image of $A$ is the whole plane, then so are the images of $A^2$ and $A^3$. If the image of $A$ is just the origin, then so is the image of $A^2$.
What remains is to study what happens when the image of $A$ is a line $\ell$. Now consider the image of $A^2$. This is equal to the image of $\ell$ under $A$. This is either all the line, or just the origin. If $A\ell$ is just the origin, then that means $A^2=0$ (one application of $A$ sends any point to $\ell$, the second application sends all those points to the origin).
If the image is all of $\ell$, then the image of $A^2$ is the same as the image of $A$, and the image of $A^3$ must again necessarily be the same $\ell$, making $A^3\neq0$.
