Proof of matrix exponential property $e^{\textbf A+\textbf B}=e^{\textbf A}e^{\textbf B}$ if $\textbf A \textbf B=\textbf B \textbf A$ I know how to prove this property with the matrix function $\pmb\Phi (t)=\exp((\textbf A+\textbf B)t)\exp(-\textbf Bt)\exp(-\textbf At)$ and its derivative $\pmb\Phi'(t)$, by proving that $\pmb\Phi'(t)=\textbf 0$ for all $t$.
It is also stated (Exercise 7.3.9, p. 539 of "Matrix Analysis and Applied Linear Algebra") that it could be proven with the function $\textbf F(t)=\exp((\textbf A+\textbf B)t)-\exp(\textbf At)\exp(\textbf Bt)$, proving that its derivative $\textbf F'(t)=\textbf 0$ so $\textbf F(t)$ is constant and since for $t=0$ it is zero so it is zero for all $t$.
I am trying to prove that $\textbf F'(t)=(\textbf A+\textbf B)\exp((\textbf A+\textbf B)t)-\textbf A\exp(\textbf At)\exp(\textbf Bt)-\textbf B\exp(\textbf At)\exp(\textbf Bt)$ is zero for all $t$ but without success. Is my derivative function correct and if it is so how to prove it is identically zero?
 A: Let's take as a starting point what you have calculated 
$\textbf F'(t)=(\textbf A+\textbf B)\exp((\textbf A+\textbf B)t)-\textbf A\exp(\textbf At)\exp(\textbf Bt)-\textbf B\exp(\textbf At)\exp(\textbf Bt)$
Then by substituting $\textbf F(t)$ to the previous expression we get
\begin{align*}
\textbf F'(t)&=(\textbf A+\textbf B)\exp((\textbf A+\textbf B)t)-\textbf A\exp(\textbf At)\exp(\textbf Bt)-\textbf B\exp(\textbf At)\exp(\textbf Bt)\\
&=(\textbf A+\textbf B)(\textbf F(t)+\exp(\textbf At)\exp(\textbf Bt))-\textbf A\exp(\textbf At)\exp(\textbf Bt)-\textbf B\exp(\textbf At)\exp(\textbf Bt)\\
&=(\textbf A+\textbf B)\textbf F(t)
\end{align*}
Now take any vector $v$ and multiply the previous equality
$$(\textbf F(t)v)'= (\textbf A+\textbf B)(\textbf F(t)v)$$ 
Which implies that 
$$\textbf F(t)v =e^{(\textbf A+\textbf B)t}\textbf F(0)v=0,\, \forall t \in \mathbb{R}$$
Therefore 
$$F(t)\equiv0$$
A: \begin{align*}
&(A+B)\exp((A+B)t) - A\exp(At)\exp(Bt)-B\exp(At)\exp(Bt)\\&=(A+B)\exp((A+B)t)-(A+B)\exp(At)\exp(Bt)\\ &=(A+B)\sum_{n=0}^{\infty}\frac{(A+B)^n t^n}{n!} - (A+B)\sum_{n=0}^\infty \frac{A^n t^n}{n!}\sum_{n=0}^\infty\frac{B^n t^n}{n!}\\
\end{align*}
But since $AB=BA$ then by standard multiplication of power series
$$\sum_{n=0}^\infty\frac{A^n t^n}{n!}\sum_{n=0}^\infty \frac{B^n t^n}{n!} = \sum_{n=0}^\infty\sum_{i=0}^n A^i B^{n-i} \frac{t^n}{i!(n-i)!} = \sum_{n=0}^\infty (A+B)^n\frac{t^n}{n!},$$
which you can prove by induction. Summarizing, the above is equivalent to
$$(A+B)\sum_{n=0}^\infty \frac{(A+B)^n t^n}{n!} - (A+B)\sum_{n=0}^\infty \frac{(A+B)^n t^n}{n!} = 0,$$
as desired.
A: $f'(t)=(A+B)e(A+B)te(-At)e(-Bt)-Ae(A+B)te(-At)e(-Bt)-Be(A+B)te(-At)e(-Bt)$
