If the matrices $A^3 = 0$, $B^3=0$ and $AB=BA$ then show this: The question: If $A$ and $B$ are square matrices of the same type such that $A^3=0$, $B^3=0$ and $AB=BA$. Show that
$$\left(I+A+\frac{A^2}{2!}\right)\left(I+B+\frac{B^2}{2!}\right)=I+(A+B)+\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$$
This is how I started. 
$$\left(I+A+\frac{A^2}{2!}\right)\left(I+B+\frac{B^2}{2!}\right)=I+(A+B)+\frac{A^2+B^2}{2!}+AB+\frac{AB^2}{2!}+\frac{A^2B}{2!}+\frac{A^2B^2}{2!2!}$$ 
I tried to get 
$$\frac{A^2+B^2}{2!}+AB+\frac{AB^2}{2!}+\frac{A^2B}{2!}+\frac{A^2B^2}{2!2!}$$
to be equal to $$\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$$ But I'm not getting there. 

I would do it without brute force:


*

*Since $A^3=0$ we have
$$
e^A=I+A+\frac{A^2}{2!}.
$$
Similar for $B$.

*The LHS of your expression is then $e^Ae^B$. For commuting matrices it is true that
$$
e^Ae^B=e^{A+B}.
$$

*Finally notice that if $A^3=B^3=0$ and $AB=BA$ then $(A+B)^5=0$ (e.g. do the binomial expansion and see that you get at least $A^3$ or $B^3$ in all terms). Thus, the RHS is precisely $e^{A+B}$.

