$\det (e^A) = e^{\text{Tr}(A)}$ without Jordan canonical form, Schur decomposition?
Let $f : \mathbb{R} \rightarrow \mathbb{C}$ be defined by $f(t) = \det e^{tA}$. This satisfies the ODE $\dot{f} = f \ \mathrm{Tr} A$, with the initial value $1$. Solving this and setting $t=1$ gives the RHS.
Added in response to comment: The map $\det : \mathrm{End}(V) \rightarrow \mathbb{C}$ is differentiable at the identity with derivative $\mathrm{Tr}$; this is a straightforward computation. We then have
$$\dot{f}(t_0) = \frac{\mathrm{d}}{\mathrm{d}t}\bigg|_{t_0} \det e^{(t-t_0)A} \det e^{t_0A}=\mathrm{Tr}\bigg(\frac{\mathrm{d}}{\mathrm{d}t}\bigg|_{t_0}(t-t_0)A\bigg) \det e^{t_0A}=f(t_0) \mathrm{Tr} A$$
for all $t_0$, using the chain rule.
If one were being super rigorous about the ODE step, one could define $g:\mathbb{R} \rightarrow \mathbb{C}$ by $g(t)=e^{t\mathrm{Tr}A}$ and consider $(f/g)^\dot{}$.
We know that
- $Tr(A) = \sum_{\forall k} \lambda_k(A)$
- $\lambda_k(e^A) = e^{\lambda_k(A)}, \forall k$
- $\det(A) = \prod_{\forall k} \lambda_k(A)$
These together with "logarithm laws": $e^{\sum a} = \prod e^{a}$ can show what we want.
It is a consequence of Jacobi formula. You can see the proof at: https://en.wikipedia.org/wiki/Jacobi%27s_formula.