The determinant of a matrix exponential is the exponential of the trace

Let $A$ be an $n \times n$ matrix over $\mathbb{C}$ . Prove that $\mathsf{det}\ \mathsf{exp}(A) = e^{\mathsf{tr}\ A}$ .

Say $P^{-1}AP = \bigoplus_i J_{k_i}(\lambda_i)$ is in Jordan canonical form, where $J_k(\lambda)$ is the Jordan block of size $k$ with eigenvalue $\lambda$ . Now $\mathsf{det}\ \mathsf{exp}(A)$ $= \mathsf{det}\ P^{-1}\mathsf{exp}(A)P$ $= \mathsf{det}\ \mathsf{exp}(P^{-1}AP)$ $= \mathsf{det}\ \mathsf{exp}(\bigoplus_i J_{k_i}(\lambda_i))$ $= \mathsf{det}\ \bigoplus_i \mathsf{exp}(J_{k_i}(\lambda_i))$ . By this previous exercise, $\mathsf{exp}(J_k(\lambda))$ is upper triangular with diagonal entries $e^\lambda$ . In particular, the determinant of this direct sum is just $\prod_i \prod_{k_i} e^{\lambda_i} = e^{\mathsf{tr}\ A}$ , since the product is over all the eigenvalues of $A$ (with multiplicity).