Inspired by my solution to 2x2 Matrix with no zero entries where $A^k=0$ - Nilpotence? I came up with this problem.
Let $A$ be a $nxn$ matrix which is not the zero matrix $0$ (in which all elements are zero).
Problems:
(i) Find an example for which $A.A\ne0$ but $A.A.A=0$.
(ii) Generalization: defining k-nilpotence for $k\ge4$ as $A^i\ne 0$ for $i=1..(k-1)$ but $A^k=0$ find examples. What can be said about the dimension of the smallest k-nilpotent matrix?
EDIT
Motivation for the second part of (ii): It can be easily shown explicitly that for a 2x2-matrix $A^3=0$ implies $A^2=0$. Hence a 3-nilpotent matrix must have at least dimension 3.
My conjecture is that this holds in general: a $k$-nilpotent matrix must have at least dimension $k$.
Hence the task is to prove of disprove the conjecture.
You figured out most of it yourself already. Let me anyway provide a little more detailed answer.
First note that a nilpotent square matrix needs to have at least a one-dimensional kernel. Otherwise it would be injective, thus surjective and no power of it could be zero. So you have finitely many possibilities for the rank of such a matrix, namely $\{n-1,\dots,0\}$. Let us, as an example, work out the case $n=3$. Consider $$A=\begin{pmatrix}0& * &*\\0&0&*\\ 0&0&0\end{pmatrix}.$$ It has rank $n-1=2$. $$A^2=\begin{pmatrix}0& 0 &*\\0&0&0\\ 0&0&0\end{pmatrix},\; A^3=\begin{pmatrix}0& 0 &0\\0&0&0\\ 0&0&0\end{pmatrix}.$$ Note that $A$ is $3$-nilpotent, $A^2$ is $2$- nilpotent and $A^3$ is $1$-nilpotent. By $k$-nilpotent I mean that $A^k=0$.
You may repeat the above construction in any dimension in the end showing that for every rank $r\in \{n-1,\dots,0\}$ you find a nilpotent matrix of that rank. Moreover, the matrix $A_r$ of rank $r$ constructed this way is $(r+1)$-nilpotent.