According to Schur decomposition the matrix $A$ can be written after a suitable change of basis as $A=D+N$ where $D$ is a diagonal matrix and $N$ is a strictly upper triangular matrix.

The formula we aim to prove

$$ \det e^A = e^{\trace A} $$

is invariant under a change of basis and thus we can carry out the computation of the exponential in any basis we choose.

By definition

\begin{equation} \label{def} e^A = \sum_{n=0}^{\infty} \frac{A^n}{n!} \end{equation} By the properties of diagonal and strictly upper triangular matrices we know that both $DN$ and $ND$ will also be strictly upper triangular matrices and so will their sum.

Thus the powers of $A$ are of the form:

\begin{eqnarray} A &=& (D+N) = D+N_1\\ A^2 &=& (D+N)(D+N) = D^2 + N_2 \\ A^3 &=& (D+N)(D^2 + N_2) = D^3 + N_3\\ &\vdots& \\ A^k &=& D^k + N_k \\ &\vdots& \end{eqnarray} where all the $N_i$ matrices are strictly upper triangular. Explicitly, $N_2 = DN_1+N_1D+N_1^2$ and by recursion $N_{n+1} = DN_n + N_n D + N_1 N_n$ .

Using equation we can write

\begin{equation} e^A = e^D + \tilde{N} \end{equation} where $\tilde{N} = \sum_{n=1}^{\infty}\frac{N_n}{n!}$ is strictly upper triangular and $e^D = \operatorname{diag}(e^{\lambda_1}, \cdots, e^{\lambda_n})$ , where $D=\operatorname{diag}(\lambda_1, \cdots,\lambda_n)$ .

$e^A$ will thus be an upper triangular matrix. Since the determinant of an upper triangular matrix is just the product of the elements in its diagonal, we can write:

\begin{equation} \label{result} \det e^A = \prod_{i=1}^{n} e^{\lambda_i} = e^{\sum_{i=1}^n \lambda_i} = e^{\trace A} \end{equation}