The derivative of the matrix exponential and the one-parameter group property

Termwise differentiation. We have $e^{tA}=\sum_k\frac{t^k}{k!}A^k$, a power series in the real variable $t$ with matrix coefficients. Its entries are real power series in $t$ with infinite radius of convergence (each entry is bounded by $\|A\|^k|t|^k/k!$ times a constant). A power series may be differentiated termwise on its interval of convergence, so
$$\frac d{dt}e^{tA}=\sum_{k\ge1}\frac{kt^{k-1}}{k!}A^k=A\sum_{k\ge1}\frac{t^{k-1}}{(k-1)!}A^{k-1}=Ae^{tA}.$$
(The matrix $A$ can be pulled out on the left or on the right: $Ae^{tA}=e^{tA}A$.)

Homomorphism. For $s,t\in\mathbb{R}$ the matrices $sA$ and $tA$ commute, so
$$e^{(s+t)A}=e^{sA}e^{tA}.$$
In particular $e^{tA}e^{-tA}=e^{0}=I$, so $e^{tA}$ is invertible with inverse $e^{-tA}$: the map takes values in $GL_n(\mathbb{R})$.

Continuity. It is differentiable, hence continuous (or: composition of the continuous maps $t\mapsto tA$ and $\exp$).

Remark. The matrix function $\Phi(t)=e^{tA}$ is the unique solution of the linear differential equation $\Phi'=A\Phi$, $\Phi(0)=I$: if $\Psi$ is another one, $\frac d{dt}\bigl(e^{-tA}\Psi(t)\bigr)=-Ae^{-tA}\Psi+e^{-tA}A\Psi=0$, so $e^{-tA}\Psi$ is constant $=I$. This uniqueness is used in the classification of one-parameter subgroups below.