Assume $A_1,A_2,...,A_n\in M_{m×m}(F)$ that satisfy the following conditions, how to prove that $A_1A_2...A_n=0$? Assume  $A_1,A_2,...,A_n\in M_{m×m}(F)$ (where $F$ is a field) such that 


*

*$A_jA_i=A_iA_j$

*$A_i^2=0, \;\;\forall 1\leq i \leq n.$ 
If $m\lt2^n$ then how to prove that $A_1A_2...A_n=0.$ 
Thanks in advance. 
 A: Lemma: Suppose $T:V\rightarrow V$ satisfies $T^2 = 0$ on an $n$ dimensional vector space $V$. Then $\mathrm{rank}(T) \le \frac{n}{2}$.
Proof: Suppose that $\mathrm{rank}(T)>\frac{n}{2}$. Then $\dim\ker(T) < \frac{n}{2}$. Therefore there must exist at least one vector in the image which is not in the kernel, for otherwise the kernel contains the image. This contradicts the fact that $T^2 = 0$. $\square$
Now let $W_i$ denote the image of the product of the first $i$ matrices. That is, $$W_i = \mathrm{Im}(A_i \cdots A_1)$$
Clearly each $W_i$ is an invariant subspace of $A_{i+1}$ for we have
$$A_{i+1}\left(A_i \cdots A_1\mathbf{x}\right) = \left(A_i \cdots A_1\right)A_{i+1}\mathbf{x}$$
by commutativity. Therefore there exists a valid restrict of each $A_{i+1}$ to $W_i$. The claim now is that $$\mathrm{rank}\left(A_{i+1}\big|_{W_i}\right)\le \frac{m}{2^{i+1}}$$
We proceed by induction.
By the previous lemma, the induction hypothesis is true for $A_1:F^m \rightarrow F^m$ with $W_0 = F^m$ since $A_1^2 = 0$. Now suppose the proposition holds for $i\ge 1$. Then consider $$A_{i+1}\big|_{W_{i}}: W_i \rightarrow W_i$$
By the induction hypothesis, $$\dim W_{i} = \mathrm{rank}\left(A_{i}\big|_{W_{i-1}}\right) \le \frac{m}{2^i}$$
Since originally $A_{i+1}^2 = 0$ on $F^m$, it follows that the restriction of the map to $W_i$ is also nilpotent. By the lemma,
$$\mathrm{rank}\left(A_{i+1}\big|_{W_i}\right) \le \frac{\dim(W_{i})}{2} \le \frac{m}{2^{i}}$$
as required. All together, the product of the matrices can be seen as $$A_n\big|_{W_{n-1}}:W_{n-1}\rightarrow W_{n-1}$$
which has rank $\frac{m}{2^n} < 1$. This means that the mapping is $0$ as required.
