Commutation of exponentials of matrices Given two $n \times n$ real matrices $A$ and $B$, prove that the following are equivalent:
(i) $\left[A,B\right]=0$
(ii) $\left[A,{\rm e}^{tB}\right] = 0,\quad$ $\forall\ t\ \in\ \mathbb{R}$
(iii) $\left[{\rm e}^{sA},{\rm e}^{tB}\right] = 0;\quad$ $\forall\ s,t\ \in\ \mathbb{R}$ 
where $\left[A,B\right] = AB - BA$ is the commutator.
First of all, this is homework, so no need for a complete answer. It is pretty easy to show that $\left(i\right)\Rightarrow\left(ii\right)$ and $\left(ii\right)\Rightarrow(iii)$. However, I have no idea for $\left(iii\right)\Rightarrow\left(i\right)$ other than explicitly writing the exponentials, and that doesn't seem to lead anywhere:
$\displaystyle{\sum_{i=0}^{\infty}\sum_{j = 0}^{\infty}
{s^{i}t^{\,j}\over i!\,j!}\,\left[A^{i},B^{\,j}\right]}.\qquad$
( I think... ) Any tips ?.

Ah, I see! $f(s,t)\equiv [e^{sA};e^{tB}] \Rightarrow \frac{\partial^2f}{\partial s \partial t}=[Ae^{sA};Be^{tB}]  \equiv 0 \Rightarrow \frac{\partial^2f}{\partial s \partial t} (0,0)=[A;B]=0$
