Prove: The determinant of the matrix exponential of A is $\displaystyle e^{tr{A}}$, i.e:
$\displaystyle \det{e^{A}}=e^{tr{A}}$
Thoughts: Look at the eigenvalues...
That's right! Look at the eigenvalues.
If $\displaystyle \lambda$ is an eigenvalue of A with eigenvector x, then $\displaystyle A^nx = \lambda^n x$ for all n, and hence $\displaystyle e^Ax = e^\lambda x$. So the eigenvalues of $\displaystyle e^A$ are the exponentials of the eigenvalues of A; and the determinant in each case is the product of the eigenvalues.
Edit. Unfortunately, that answer was incomplete, because it does not deal with multiplicity. You need the matrices A and $\displaystyle e^A$ not only to have the same eigenvalues, but to have them with the same multiplicity. One way to do this is to see that the Jordan normal form of $\displaystyle e^A$ has the exponentials of the eigenvalues of A as its diagonal elements. In the answer I gave above, I was trying to avoid having to use the Jordan normal form, but maybe it's unavoidable.