For any matrix $A$ the series $$\bex \exp A=I+A+\frac{A^2}{2!}+\cdots+\frac{A^n}{n!}+\cdots \eex$$ converges. This is called the exponential of $A$. The matrix $A$ is always invertible and $$\bex (\exp A)^{-1}=\exp(-A). \eex$$ Conversely, every invertible matrix can be expressed as the exponential of some matrix. Every unitary matrix can be expressed as the exponential of a skew-Hermitian matrix.
Solution.
(1). $$\bex \exp A=\sum_{n=0}^\infty \frac{A^n}{n!} \eex$$ follows from the fact that $$\bex \sum_{n=0}^\infty \frac{\sen{A}^2}{n!}=\exp \sen{A}<\infty \eex$$ and the completeness of $\M(n)$.
(2). By taking limits in $$\beex \bea &\quad\sex{\sum_{k=0}^n\frac{A^k}{k!}} \cdot \sex{\sum_{l=0}^n \frac{B^l}{l!}}\quad\sex{AB=BA}\\ &=\sum_{k,l=0}^n \frac{A^kB^l}{k!l!}\\ &=\sum_{s=0}^{2n} \frac{1}{s!}\sum_{k+l=s}\frac{s!}{k!(s-k)!}A^kB^{s-k}\\ &=\sum_{s=0}^{2n}\frac{1}{s!}(A+B)^s, \eea \eeex$$ we have $$\bex \exp(A)\cdot \exp (B)=\exp(A+B). \eex$$ Taking $B=-A$, we see readily that $$\bex \exp(A)\cdot \exp(-A)=I. \eex$$
(3). For invertible matrix $A$, by theJordan canonical decomposition, there exists an unitary $U$ such that $$\bex A=U\diag(J_1,\cdots,J_s)U^*, \eex$$ with the diagonals $\lm_i$ of $J_i$ is not equal to zero. We only need to show that $J_i$ is the exponential of some matrix. In fact, set $\mu_i\in\bbC$ satisfy $e^{\mu_i}=\lm_i$ and $$\bex \vLm_i=\diag(\mu_i,\cdots,\mu_i), \eex$$ then its exponential $$\bex \exp \vLm_i=\diag(\lm_i,\cdots,\lm_i) \eex$$ has the same eigenvalues of $J_i$. Hence, they are similar, and there exists some invertible matrix $P_i$ such that $$\bex J_i=P_i^{-1}\exp \vLm_i P_i=\exp [P_i^{-1}\vLm_iP_i]. \eex$$
(4). For $U\in \U(n)$, $$\bex U=\exp B\ra I=U^*U=\exp (B^*)\cdot \exp (B)=\exp(B^*+B)\ra B^*=-B. \eex$$
