Let $(A,\mathfrak{m})$ be a local ring. Let us call $\mathfrak{m}$ almost nilpotent if for every sequence $a_1,a_2,\dotsc$ in $\mathfrak{m}$ there is some $n \geq 1$ such that $a_1 \cdot \dotsc \cdot a_n = 0$. Let $M$ be an $A$-module with $M = \mathfrak{m} M$. I would like to prove $M = 0$.
This is an exercise in Lenstra's Galois Theory for Schemes I've been struggling with quite some time. If $M$ is finitely generated, it is trivial (Nakayama). Now let's say $M$ is countably generated, by $m_1,m_2,m_3,\dotsc$. By elimination, we may then assume that $m_i \in \langle m_{i+1},m_{i+2},\dotsc \rangle$. If we had $m_i \in \langle m_{i+1} \rangle$, it would be easy to conclude: Choose $a_i \in \mathfrak{m}$ with $m_i = a_i m_{i+1}$. Now apply the assumption to the sequence $(a_i)$, this shows $m_1 = 0$. Since we could choose $m_1$ arbitrary, $M=0$. However, in general case, the equations become quite horrible and sums, not just products, are involved. If you draw a tree representing the linear combinations, I know that every path must end in a zero eventually, but not that the whole tree must end eventually. My gut feeling is that there may be universal counterexamples ...