For any matrix $A$ the series $$ex exp A=I+A+frac{A^2}{2!}+cdots+frac{A^n}{n!}+cdots eex$$ converges. This is called the exponential of $A$. The matrix $A$ is always invertible and $$ex (exp A)^{-1}=exp(-A). eex$$ 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). $$ex exp A=sum_{n=0}^infty frac{A^n}{n!} eex$$ follows from the fact that $$ex sum_{n=0}^infty frac{sen{A}^2}{n!}=exp sen{A}<infty eex$$ and the completeness of $M(n)$.
(2). By taking limits in $$eex ea &quadsex{sum_{k=0}^nfrac{A^k}{k!}} cdot sex{sum_{l=0}^n frac{B^l}{l!}}quadsex{AB=BA}\ &=sum_{k,l=0}^n frac{A^kB^l}{k!l!}\ &=sum_{s=0}^{2n} frac{1}{s!}sum_{k+l=s}frac{s!}{k!(s-k)!}A^kB^{s-k}\ &=sum_{s=0}^{2n}frac{1}{s!}(A+B)^s, eea eeex$$ we have $$ex exp(A)cdot exp (B)=exp(A+B). eex$$ Taking $B=-A$, we see readily that $$ex exp(A)cdot exp(-A)=I. eex$$
(3). For invertible matrix $A$, by theJordan canonical decomposition, there exists an unitary $U$ such that $$ex A=Udiag(J_1,cdots,J_s)U^*, eex$$ 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 $mu_iinbC$ satisfy $e^{mu_i}=lm_i$ and $$ex vLm_i=diag(mu_i,cdots,mu_i), eex$$ then its exponential $$ex exp vLm_i=diag(lm_i,cdots,lm_i) eex$$ has the same eigenvalues of $J_i$. Hence, they are similar, and there exists some invertible matrix $P_i$ such that $$ex J_i=P_i^{-1}exp vLm_i P_i=exp [P_i^{-1}vLm_iP_i]. eex$$
(4). For $Uin U(n)$, $$ex U=exp B a I=U^*U=exp (B^*)cdot exp (B)=exp(B^*+B) a B^*=-B. eex$$