The following is an excerpt from a book I'm reading:
The following property of the matrix exponential can readily be established by a variety of methods-the easiest perhaps being the use of the series definition (3.4)- \begin{equation} \tag{3.8} e^{A(t_1 + t_2)} = e^{At_1}e^{At_2} \end{equation} for any $t_1$ and $t_2$. From this property, it follows that \begin{equation} \tag{3.9} \left(e^{A\tau}\right)^{-1} = e^{-A\tau} \end{equation}
How does $(3.9)$ follow from $(3.8)$?
Note that the matrix $B$ is the inverse of the (square) matrix $C$ if and only if $CB = I$, and its the notational convention to call the matrix $B$ which solves this $C^{-1}$.
So for the matrix $C = e^{A\tau}$, its inverse $(e^{A\tau})^{-1}$ is the matrix $B$ such that $$e^{A\tau}B = I.$$
From (3.8) we have $$ e^{A\tau}e^{-A\tau}=e^{A(\tau + (-\tau))} = e^{A0} = I$$ which shows that the matrix $B = e^{-A\tau}$ is the inverse of $e^{A\tau}$, that is $(e^{A\tau})^{-1} = e^{-A\tau}$.