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 ... 

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.
