Stuck on matrix exponential problem I want to show that the property $e^{tA}e^{tB}=e^{t(A+B)}$ implies that $AB=BA$. Here $A,B$ denotes matrices and $t\in \mathbb{R}$.

Im stuck, have tried to expand both sides with their taylor series but not sure if this is the way to go.

Could anyone give me a hint?
Differentiate 2 times with respect to time $t$ the identity 
$$e^{tA}e^{tB}=e^{t(A+B)}$$
and then set $t=0$.
This is not what has been asked but to show the identity is valid if $A$ and $B$ commutes, you can...

Define
$$
f(t) = e^{t(A+B)}e^{-tB}e^{-tA}
$$

Then,
$$
f'(t) = (A+B)e^{t(A+B)}e^{-tB}e^{-tA} - e^{t(A+B)}Be^{-tB}e^{-tA}
- e^{t(A+B)}e^{-tB}Ae^{-tA}
$$

if $A$ and $B$ commutes, we can factor out A and B from the last two.
$$
f'(t) = (A+B) f(t) - A f(t) - B f(t) = 0
$$

Therefore $f(t)$ is a constant, you can show that it's identity.