Why is $\det(e^X)=e^{\operatorname{tr}(X)}$?

Question

I've seen on Wikipedia that for a complex matrix $X$, $\det(e^X)=e^{\operatorname{tr}(X)}$.

It is clearly true for a diagonal matrix. What about other matrices ?

The series-based definition of exp is useless here.

Answer 1

A alternative to doing this by normal forms which perhaps assumes more but is much more natural to me is (as suggested in the comment on $\det(\exp X)=e^{\mathrm{Tr}\, X}$ for 2 dimensional matrices) to note that it clearly holds for diagonalizable matrices (see the duplicate How to prove $\det(e^A) = e^{\operatorname{tr}(A)}$? ...), and by

1. the continuity of $\det, \mathrm{tr}$ and $\exp$
2. the density of diagonalizable matrices in the space of all complex matrices (Diagonalizable matrices with complex values are dense in set of $n\times n$ complex matrices.)

we have the result more generally for all matrices.

Answer 2

If $X$ is upper triangular, then this is clear.

If the claim holds for a matrix $Y$, then it holds for any $X$ similar to $Y$.

By Jordan decomposition, each $X$ is similar to an upper triangular matrix $Y$ (of a special form, but never mind).

Thus, the claim holds for all $X$.