<s>[INST] 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. 
 [/INST] You can simply match the terms of second order, third order and fourth order to the corresponding terms in the answer:
For second order terms: by multiplication $(A+B)^2=A^2+AB+BA+B^2$ and using $AB=BA$ we see that 
$$
\frac{(A+B)^2}{2!} = \frac{A^2+B^2}{2} + AB
$$
For third order terms we have by multiplication $(A+B)^3 = A^3 + (A^2B+ABA+B^2A) + (AB^2+BAB+B^2A) + B^3$.  By $A^3=B^3=0$ this becomes
$(A+B)^3=(A^2B+ABA+B^2A) + (AB^2+BAB+B^2A)$ and using $AB=BA$ gives 
$(A+B)^3=3(A^2B+AB^2)$.  The commutativity property simply allows us to reorder the terms in the multiplications.  Therefore
$$
\frac{(A+B)^3}{3!}=\frac{A^2B+B^2A}{2!}
$$.
Finally, for the fourth order term we note that by commutativity we have
$$
(A+B)^4 = A^4+4A^3B+6A^2B^2+4AB^3+B^4
$$
Since $A^3=B^3=0$ only the middle term remains.  Hence
$$
\frac{(A+B)^4}{4!} = \frac{A^2B^2}{2!2!}.
$$
</s>