The identity det(A) = exp(Tr(ln(A)) for A general I understand the proof of the identity in the title for $A$ Hermitian. One uses that any Hermitian matrix can be diagonalized as $A = X \Lambda X^{-1}$, such that
$$
\det{A} = \prod_i \lambda_i,
$$
and we have
$$
\exp(Tr(\log(A)) = \exp(Tr(X\log\Lambda X^ {-1}) = \exp(\sum_i\log(\lambda_i)) = \prod_i \lambda_i.
$$
However, is it possible to show the identity for $A$ not Hermitian? My motivation for this question is that in physics the identity is often used without it being clear that $A$ is Hermitian.
 A: 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 $y(0) = 1$.
A: Here is an elegant proof with minimal assumptions (only non-singular and square):
\begin{align}
\rm{Det}(e^{B})  &= \displaystyle \lim_{N \to \infty} \rm{Det}\Big(e^{B/N}\Big)^{N} \\ 
&= \displaystyle \lim_{N \to \infty} \rm{Det} \Big(1 + \frac{B}{N}\Big)^{N}  \\ 
&= \displaystyle \lim_{N \to \infty} \Big(1 + \frac{\rm{Tr}B}{N}\Big)^{N} \\ 
&= e^{\rm{Tr}B} 
\end{align}
Now take log on both sides and put $ e^{B} = A$ to get the result. 
