The identity det(A) = exp(Tr(ln(A)) for A general

If $\ln(A) = B$, the identity says

$$ \det(\exp(B)) = \exp(\text{Tr}(B)) $$

which is more usual form for this identity, true for all $n \times n$ matrices $B$ over $\mathbb C$ (avoiding questions about whether $\ln(A)$ is defined, and which of the possible logarithms to use).

One way to do this is to show it first for diagonalizable matrices $B$, then use the fact that diagonalizable matrices are dense and both sides of the equation are continuous functions of $B$.

A second way is to use Jordan canonical form.

A third way is to note that both $\det(\exp(tB))$ and $\exp(t \text{Tr}(B))$ satisfy the differential equation $y' = \text{Tr}(B) y$ with initial value $y(0) = 1$.