TITLE: Proving that the matrix exponential map is surjective onto the general linear group
QUESTION [18 upvotes]: Let $M_n(\mathbb{F})$ be the set of all $n\times n$ with entries in $\mathbb{F}$ and let $\exp:M_n(\mathbb{C})\to M_n(\mathbb{C})$ be defined by 
$$ \exp(A)=\sum_{k=0}^{\infty}\frac{A^k}{k!},$$ for all $A\in M_n(\mathbb{C}).$
I want to prove that $\exp$ is a surjective map from $M_n(\mathbb{C})$ to $GL(n,\mathbb{C})=\left\{A\in M_n(\mathbb{C})\,\middle| \det(A)\neq0\right\}$, how do I go about that? 
I mean saying that $\exp:M_n(\mathbb{R})\to GL(n,\mathbb{R})$ is an analogous to saying $\exp:\mathbb{R}\to \mathbb{R}_{>0}$ and this is also pretty intuitive, since, in analogy with the case of numbers, $A^0=I\;\forall A$, so
$\exp(0)=I+0+\frac{0^2}{2!}+\dots=I$, so even for $A=0$ we get $\det\left(\exp(A)\right)\neq0$ and so because of the first term we can never get a zero determinant. But I have no idea how to prove the subjectiveness. 
Thanks in advance.

REPLY [3 votes]: I am thinking about a simple solution (perhaps simple-minded). For a diagonalizable matrix, it can be obtained by exponentiation, because for each eigenvalue $w\neq 0$, $e^z=w$ has a solution in $\mathbb{C}$. Diagonalizable matrices are dense in $GL(n,\mathbb{C})$, so for any $M \in GL(n,\mathbb{C})$, there exists a sequence of diagonalizable matrix $M_k$ that converges to $M$, and each can be written as $M_k = e^{A_k}$. Convergence imposes bound on the norm of $M_k$ hence on $A_k$, so we have a subsequence of $A_k$ converging to $A$. By continuity of the exponential map we should have $e^A=M$.