The determinant of an exponential: det e^A = e^(tr A)

Idea. Reduce to triangular matrices, where both sides can be computed on the diagonal.

Step 1: conjugation. For invertible $P$ and any $X$, $(PXP^{-1})^k=PX^kP^{-1}$, so
$$e^{PXP^{-1}}=Pe^XP^{-1}.$$
(Termwise, and $X\mapsto PXP^{-1}$ is linear and continuous.)

Step 2: triangularisation. Over $\mathbb{C}$ (Schur or Jordan form) there are an invertible $P\in GL_n(\mathbb{C})$ and an upper triangular $T$ with $A=PTP^{-1}$. The same series definition works in $M_n(\mathbb{C})$, and the identity of step 1 still holds. Hence $e^A=Pe^TP^{-1}$.

Step 3: triangular matrices. Let $T$ be upper triangular with diagonal entries $t_{11},\dots,t_{nn}$. Powers of $T$ are upper triangular and the diagonal of $T^k$ is $(t_{11}^k,\dots,t_{nn}^k)$. So $e^T$ is upper triangular with diagonal entries $\sum_kt_{ii}^k/k!=e^{t_{ii}}$.

Step 4: conclusion. The determinant of a triangular matrix is the product of its diagonal entries, so
$$\det e^A=\det(Pe^TP^{-1})=\det e^T=\prod_ie^{t_{ii}}=e^{\sum t_{ii}}=e^{\operatorname{tr}T}=e^{\operatorname{tr}A},$$
because the trace is invariant under conjugation.

Consequence. For real $A$, $e^A$ is real and $\det e^A=e^{\operatorname{tr}A}>0$. Hence $\exp(M_n(\mathbb{R}))\subset GL_n^+(\mathbb{R})$ and in particular $\exp(\mathfrak{sl}_n)\subset SL_n$.

Check. For $A=\theta J$ we have $\operatorname{tr}A=0$ and $\det R_\theta=1$. For $A=\operatorname{diag}(a,b)$, $\det e^A=e^{a+b}$.

Remark. An alternative proof: $f(t)=\det e^{tA}$ satisfies $f'(t)=\operatorname{tr}(A)f(t)$ (Jacobi's formula), so $f(t)=e^{t\operatorname{tr}A}$.