Let $A$ be an $n \times n$ matrix over $\mathbb{C}$ or $\mathbb{R}$. Does $\det(e^A) = e^{\mathrm{tr}(A)}$ always hold?

For an alternative argument, you can compute $f'(t)$, with $f(t)=\det e^{tA}$. Since $f(t_0+s)=f(t_0)f(s)$, it suffices to do this at $t=0$. Then $e^{tA}=1+tA+O(t^2)$, so $$ \det e^{tA}=1+t\,\textrm{tr}\, A +O(t^2) , $$ because the only way to obtain a contribution linear in $t$ is to multiply one of the $ta_{jj}$ on the diagonal with the $1$'s in the other diagonal entries. This says that $f'(0)=\textrm{tr}\, A$.

So $f$ solves the IVP $f'=(\textrm{tr}\, A)f$, $f(0)=1$. Thus $f(t)=e^{t\,\textrm{tr} A}$, as claimed.

An Argument from Lie Theory

Using the Jordan-Chevalley Decomposition Theorem (for the Lie algebra $\mathfrak{sl}_n(\mathbb{C})$, which consists of traceless $n$-by-$n$ matrices), $A=t\,I+S+N$ where $t:=\frac{\text{trace}(A)}{n}$, $S$ is semisimple (i.e., diagonalizable over $\mathbb{C}$) of trace $0$ , and $N$ is nilpotent such that $SN=NS$. Then, because $t\,I$, $S$, and $N$ commute, $\exp(A)=\exp(t\,I)\,\exp(S)\,\exp(N)$. That is, $$\det\big(\exp(A)\big)=\det\big(\exp(t\,I)\big)\cdot\det\big(\exp(S)\big)\cdot\det\big(\exp(N)\big)\,.$$ Obviously, the first factor is $\det\big(\exp(t\,I)\big)=\big(\exp(t)\big)^n=\exp(nt)=\exp\big(\text{trace}(A)\big)$. The last factor is $1$ because $\exp(N)=I+M$ for some nilpotent matrix $M$ (whence the eigenvalues of $M$ are all $0$). The middle factor is $1$ because $S$ is diagonalizable as $TDT^{-1}$ for some invertible matrix $T$ and for some diagonal traceless matrix $D$. Then, $$\det\big(\exp(S)\big)=\det\big(T\exp(D)T^{-1}\big)=\det\big(\exp(D)\big)=\exp\big(\text{trace}(D)\big)=1\,.$$ Consequently, $\det\big(\exp(A)\big)=\exp\big(\text{trace}(A)\big)$.