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 ?.
 A: The easiest way that I see is to do it in two steps, proving $(iii) \Rightarrow (ii) \Rightarrow (i)$.
To prove $(iii) \Rightarrow (ii)$, differentiate $f(s) = [e^{sA},e^{tB}] \equiv 0$. The proof of $(ii) \Rightarrow (i)$ is quite similar.
A: Ah, I see! $f(s,t)\equiv [e^{sA};e^{tB}] \Rightarrow \frac{\partial^2f}{\partial s \partial t}=[Ae^{sA};Be^{tB}]  \equiv 0 \Rightarrow \frac{\partial^2f}{\partial s \partial t} (0,0)=[A;B]=0$
A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
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 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
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\begin{align}
&\partiald{\bracks{{\rm e}^{sA},{\rm e}^{tB}}}{s}
= \bracks{\expo{sA}A,\expo{tB}} = 0\,,\qquad
{\partial^{2}\bracks{{\rm e}^{sA},{\rm e}^{tB}} \over \partial t\,\partial s}
= \bracks{\expo{sA}A,\expo{tB}B} = 0
\end{align}
In particular, with $t = s = 0$: $\bracks{A,B} = 0$.
