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?
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)&amp;=(\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)\\
&amp;=(\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)\\
&amp;=(\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 
$$F(t)\equiv0$$
$f'(t)=(A+B)e(A+B)te(-At)e(-Bt)-Ae(A+B)te(-At)e(-Bt)-Be(A+B)te(-At)e(-Bt)$