Show that:
\begin{equation} \dv t e^{tA} = e^{tA}A \end{equation}
We can apply the result we shown in eigenvalue:
\begin{equation} \dv t \left(e^{tA}\right) = \dv t \left(I + \sum_{k=1}^{\infty} \frac{t^{k}}{k!}A^{k}\right) = \left(\sum_{k=1}^{\infty }\frac{1}{k!}kt^{k-1}A^{k-1}\right)A \end{equation}
We do this separation because k=0 would’t make sense to raise A (k-1=-1) to as we are unsure about the invertability of A. Obviously \frac{1}{k!}k = \frac{1}{(k-1)!}. Therefore, we can shift our index back yet again:
\begin{equation} \left(\sum_{k=1}^{\infty }\frac{1}{k!}kt^{k-1}A^{k-1}\right)A = \left(\sum_{j=0}^{\infty }\frac{1}{j!}t^{j}A^{j}\right)A \end{equation}
Awesome. So now we have the taylor series in e^{tA} back, times A. So therefore:
\begin{equation} \left(\sum_{j=0}^{\infty }\frac{1}{j!}t^{j}A^{j}\right)A = e^{tA}A \end{equation}
Be forewarned:
\begin{equation} e^{A}e^{B} \neq e^{A+B} \end{equation}
mostly because matrix multiplication is not commutative..