Convergence (Property 1). For any matrix norm, $\|A^k t^k/k!\| \leq \|A\|^k |t|^k/k!$. The series $\sum \|A\|^k|t|^k/k! = e^{\|A\||t|}$ converges, so the matrix series converges absolutely.

Initial condition and ODE (Property 2). At $t = 0$: $e^{A\cdot 0} = I + 0 + \cdots = I$. For the derivative: term-by-term differentiation (justified by uniform convergence on compact sets) gives

$\frac{d}{dt}e^{At} = \sum_{k=1}^\infty \frac{kA^k t^{k-1}}{k!} = A\sum_{k=1}^\infty \frac{A^{k-1}t^{k-1}}{(k-1)!} = A\sum_{j=0}^\infty \frac{(At)^j}{j!} = Ae^{At}.$

Group property (Property 3). Since $A$ commutes with itself, the Cauchy product of the series for $e^{At}$ and $e^{As}$ yields $e^{A(t+s)}$:

$e^{At}e^{As} = \left(\sum_{j=0}^\infty\frac{A^jt^j}{j!}\right)\left(\sum_{k=0}^\infty\frac{A^ks^k}{k!}\right) = \sum_{n=0}^\infty\sum_{j+k=n}\frac{A^n t^j s^k}{j!k!} = \sum_{n=0}^\infty\frac{A^n(t+s)^n}{n!} = e^{A(t+s)}.$

Invertibility (Property 4). From Property 3 with $s = -t$: $e^{At}e^{-At} = e^{A\cdot 0} = I$.

Determinant (Property 5). Let $\phi(t) = \det(e^{At})$. Then $\phi(0) = 1$ and $\phi(t+s) = \phi(t)\phi(s)$ (since $\det(AB) = \det A\det B$). Differentiating: $\phi'(0) = \text{tr}(A)$ (by the formula for the derivative of the determinant). The only continuous solution to $\phi(t+s) = \phi(t)\phi(s)$ with $\phi'(0) = \text{tr}(A)$ is $\phi(t) = e^{\text{tr}(A)t}$. $\blacksquare$