Mutually commuting matrix with $A_i^2=0$ Let $A_1,\dots,A_n$ be mutually commuting $m\times m$ matrices such that $A_i^2=0$ for all $1\le i \le n$. If $m<2^n$, prove that $A_1 A_2\cdots A_n=0$

Since $A_i^2=0$ So $\operatorname{Im}(A)\subset \ker(A) \implies \dim(\ker(A))\ge \frac m 2$ I think we have to use it to prove ..but don't know how go through ... 
 A: Let $x\in K^m$.
Then $A_1x\in Im(A_1)$, vector space of dimension $\leq m/2$. 
In the sequel, $\tilde{U}$ denotes a restriction of $U$.
$\tilde{A_2}:Im(A_1)\rightarrow Im(A_1)$ is $2$-nilpotent. Then $A_2A_1x\in Im(\tilde{A_2})$, 
vector space of dimension $\leq dim(Im(A_1))/2\leq m/2^2$, and so on  $\tilde{A_3}:Im(\tilde{A_2})\rightarrow Im(\tilde{A_2})$ is $2$-nilpotent,....
$\tilde{A_n}:Im(\tilde{A_{n-1}})\rightarrow Im(\tilde{A_{n-1}})$ 
 is $2$-nilpotent. Then $A_n\cdots A_1x\in Im(\tilde{A_n})$, vector space of dimension $\leq dim(Im(\tilde{A_{n-1}}))/2\leq m/2^n<1$.
Finally $Im(\tilde{A_n})=\{0\}$ and we are done.
A: Edit.
The answer below seems to be only the partial answer for the particular form of nilpotent matrices.

Mutually commuting means that all matrices can be represented simultaneously as, for example, upper triangular.   
Of course they are all nilpotent and (here assumption for the form of matrices) the first non-zero "mini-diagonal" of this matrix is shifted from the main diagonal at least rounded $m/2$.
(do yourself analysis for dimension $m$ even and odd)
For example matrix
$\begin{bmatrix}
0\color{red}\rightarrow & 0\color{red}\rightarrow & 0\color{red}\rightarrow & \color{red}1 & 3 \\
0 & 0\color{red}\rightarrow & 0\color{red}\rightarrow & 0\color{red}\rightarrow & \color{red}2 \\
0 & 0 & 0 & 0 & 0  \\
0 & 0 & 0 & 0 & 0  \\
0 & 0 & 0 & 0 & 0  \\
\end{bmatrix}$
has first minidiagonal  (red) shifted by $3$ positions from the main diagonal.
In multiplication values of shifts for such matrices are added.
I mean that if $A=A_1A_2$ then shift($A$)=shift($A_1$)+shift($A_2$).
In this case $A^2$ should have to have shift $6>5$ what means that $A^2=0$.
(If shift were by $2$ with $A^2$ result shift would be $ 4$ what is too little to make matrix zero in general case.)  
Multiplication $A_1A_2 \dots A_n$ means that   the result matrix has shift at least $nm/2$ what is 
fully fulfilled if $m<2^n$.
Look at some examples.
For $n=2$ we have limit for $m$ equal 3. Result the least shift $3$.  
For $n=3$ we have limit for $m$ equal 7.  Result shift $12$ 
For $n=4$ we have limit for $m$ equal 15.  Result shift $30$,
Always result shift is bigger then $m$ what means that we get zero matrix.
