Derivative of the matrix exponential

Power series.
$$e^{tA} = \sum_{n=0}^{\infty} \frac{(tA)^n}{n!} = \sum_{n=0}^{\infty} \frac{t^n A^n}{n!}.$$

Differentiation.
$$\frac{d}{dt} e^{tA} = \sum_{n=1}^{\infty} \frac{n t^{n-1} A^n}{n!} = \sum_{n=1}^{\infty} \frac{t^{n-1} A^n}{(n-1)!} = A \sum_{n=1}^{\infty} \frac{t^{n-1} A^{n-1}}{(n-1)!} = A e^{tA}.$$
Since $A$ commutes with $e^{tA}$, we also have $e^{tA} A$.

Remark. This property is the key to solving linear systems $X' = AX$ via $X(t) = e^{tA}X(0)$.