How to show that the matrix exponential is invertible for non-diagonalizable matrix A, I have shown the easy case, when A is diagonalizable.
But I am stuck on the case when A is not.
So, I put A in its Jordan canonical form.  Then say A = $SJS^{-1}$.
Then $e^A = Se^JS^{-1}$, 
where $e^J$ is an upper triangular matrix with the Jordan blocks exponentiated.  
Now for each Jordan block, I have that 
$$e^{J_i} = e^{\lambda_i I + N}$$
where N is elementary nilpotent.  But since scalar matrices commute with all matrices, it commutes with N.
Then 
$$e^{J_i} = e^{\lambda_i I + N} = e^{\lambda_i}e^{N}$$
where $e^{\lambda_i}$ is a diagonal matrix with non-zero diagonal, hence it is invertible / has non-zero determinant.  What can I do with the $e^N$ factor?  I know that it has a finite expansion, since it is nilpotent:
$$e^N = I + N + ... + \frac {N^k}{k!}$$
I'm not sure where to go from here...
Thanks,
 A: The eigenvalues $\lambda_i$ become $e^{\lambda_i}$ after matrix exponential and the exponential function has no zeros. Therefore any matrix over an algebraically closed field will have a matrix exponential full of non-zero eigenvalues and we don't even need to use Jordan.
To clarify, we just need to use definition of eigenvalue and the fact that degeneracy (non-invertibility) is the same as a one or more eigenvalues equal to 0. The eigenvalues are the roots to the characteristic polynomial which if matrix is under algebraically closed field must have as many of them as it's size. So there will always be $n$ eigenvalues regardless of diagonalization or jordan form and at least 1 of those needs to be 0 to make the matrix degenerate.

Important note: This only works when working over any field which the exponential function has no zeroes. I don't know if there may exist any fields where the exponential function can actually give 0.
A: As $e^N$ has only $1$s on its diagonal (the $\sum_{k\ge 1} N^k/k!$ part is strictly upper diagonal), $e^{J_i}$ has only $e^{\lambda_i}$ on its diagonal. We therefore have
$$ \det(e^{J_i}) = e^{k_i\lambda_i} $$
where $k_i$ is the size of $J_i$, this gives 
$$ \det(e^A) = \det S^{-1}\prod_i \det(e^{J_i}) \det S 
   = e^{\sum_i k_i \lambda_i} = e^{\mathrm{tr}\, A} $$
As $0 \not\in \exp(\mathbf C)$, we have that $\det(e^A) = e^{\text{something}} \ne 0$.
A: The simpler way to shows that $e^A$ is invertible is to note that, for commuting matrices $A,B$ we have, from the definition of the exponential, $e^{A+B}=e^Ae^B$. So, since $A$ and $-A$ commute, we have:
$$
e^Ae^{-A}=e^{A-A}=e^O=I
$$
Also, using Jacobi's formula (see here) we can find:
$$
\det(e^A)=e^{Tr\;(A)}
$$
that confirms the invertibility of $e^A$.
