TITLE: Prove $\mathrm{exp}(\mathrm{Tr}(A)) = \det(\mathrm{exp}(A))$ via maximal tori
QUESTION [5 upvotes]: This is an exercise in Kris Tapp's little AMS booklet Matrix Groups for Undergraduates. I have reviewed the proof given earlier in the text but that proof relies on properties of exp and other aspects of commuting matrices, the definition of the derivative and other simpler observations.  Ah! Commuting matrices must be one  key here but I am afraid I haven't made enough of  the connections. My feeling is that conjugation will be helpful as well, since all maximal tori are conjugates of each other.
I believe I have all the parts of the jigsaw puzzle but have yet to fit the appropriate pieces. Maximal Tori is the name of the Chapter (#9) and they are, hence, a NEW topic, which again points toward their properties to be the glue to  hold the pieces together.
Any suggestions?

REPLY [2 votes]: The result clearly holds for $A$ diagonal, and thus it holds for $A$ diagonalizable by the fact that $\operatorname{tr}$ and $\det$ are both invariant under conjugation. But diagonalizable matrices are dense in $M_n(\mathbb{C})$ (in the usual metric), so the result holds for arbitrary $A$.