I know how to show that $\det(e^A)=e^{{\rm tr}(A)}$ by elementary methods using the eigenvalues of $e^A$ (that is, if $\lambda$ is an eigenvalue of $A$ then $e^\lambda$ is an eigenvalue of $e^A$).
However I found an exercise that want that the identity be proved in a very different way (more analytical). It says:
Let $f(t):=\det(e^{tA})-e^{{\rm tr}(tA)}$. Now, using the derivative of the determinant, show that $f'=0$.
The derivative of the determinant is stated as:
Let $a_1,\ldots,a_m\in C^1(X,\Bbb K^m)$, where $X\subset\Bbb K$ is open, then $\det[a_1,\ldots,a_m]\in C^1(X,\Bbb K)$ and $$(\det[a_1,\ldots,a_m])'=\sum_{k=1}^m\det[a_1,\ldots,a_{k-1},a_k',a_{k+1},\ldots,a_m]$$
Here $\Bbb K\in\{\Bbb R,\Bbb C\}$ and the derivative is clear as the derivative of a $m$-linear function for the $a_k$ as vector columns (or rows) of a $m\times m$ matrix.
Trying to use this I can see that
$$f'(t)=\partial\det(e^{tA})e^{tA}A-{\rm tr}(A)e^{t\,{\rm tr}(A)}$$
But it is far to be clear how the identity of the question can be proved from here, that is, how to prove that $f'=0$. Anyone knows how to handle this proof? Thank you.
Using the identity you gave, the computation is $$\det'(e^{tA})e^{tA}A=\sum_{k=1}^m \det(e^{tA}e_1,..,e^{tA}Ae_k,..,e^{tA}e_m),$$ where $(e_1,..,e_m)$ is the canonical basis. So we get $$\det'(e^{tA})e^{tA}A=\det(e^{tA}) \left( \sum_{k=1}^m \det(e_1,..,Ae_k,..,e_m) \right)=\det(e^{tA}) tr(A).$$ Plugin this into the expression for $f'$ gives $$f'(t)=tr(A) f(t),$$ and this differential equation is now easy to solve: $f(t)=e^{tr(A)t}f(0)=0.$
(There may be a simpler way...)