TITLE: How to prove $\det(e^A) = e^{\operatorname{tr}(A)}$?
QUESTION [28 upvotes]: Prove $$\det(e^A) = e^{\operatorname{tr}(A)}$$ for all matrices $A \in \mathbb{C}_{n×n}$.

REPLY [17 votes]: Let $f(t)= \det(e^{tA})$. Then $f'(t)=D \det(e^{tA}) \cdot Ae^{tA}=\text{tr} \left(^t \text{com}(e^{tA})Ae^{tA} \right)$. But $A$ and $e^{tA}$ commute, and $^t\text{com}(e^{tA})e^{tA}=\det(e^{tA}) \operatorname{I}_n$. Therefore, $f'(t)=\text{tr}(A)f(t)$ and $f(0)=1$, hence $f(t)=e^{\text{tr}(A)t}$. For $t=1$, $\det(e^{A})= e^{\text{tr}(A)}$.