Prove $\exp(\mathrm{Tr}(X))=\det(\exp(X))$ 
Show that $\exp(\mathrm{Tr}(X))=\det(\exp(X))$ where $X$ is a matrix using the concept of the Jordan normal form

I realised this formula by considering that: $\det(\exp(X))=\exp(\lambda_1) \times\cdots \times \exp(\lambda_n)=\exp(\lambda_1 + \cdots + \lambda_n)=\exp(\mathrm{Tr}(X))$ 
$\lambda_i$ are eigenvalues of $X$
I was unsure how to prove it using the Jordan normal form - could you help me please?
 A: If $X$ is similar to $Y$ with $X = P^{-1}YP$ then
$$ \operatorname{Tr}(X) = \operatorname{Tr}(Y) \implies \exp(\operatorname{Tr}(X)) = \exp(\operatorname{Tr}(Y)), \\
\exp(X) = P^{-1}\exp(Y)P \implies \det(\exp(X)) = \det(\exp(Y))
$$
so we can assume that $X \in M_n(\mathbb{C})$ is in Jordan canonical form. In particular, $X$ is upper triangular with the eigenvalues $\lambda_1, \dots, \lambda_n$ (counted with multiplicities) on the diagonal. Then, by the definition of $\exp$, the matrix $\exp(X)$ is also upper triangular with the values $e^{\lambda_1}, \dots, e^{\lambda_n}$ on the diagonal and so 
$$ \det(\exp(X)) = e^{\lambda_1} \dots e^{\lambda_n} = e^{\lambda_1 + \dots + \lambda_n} = \exp{\operatorname{Tr(X)}}. $$
A: Here is another approach using differential equations:
Let $Y$ solve $\dot{Y} = XY$ subject to $Y(0) = I$. Note that the unique solution is given by
$Y(t) = e^{tX}$.
Let $d(t) = \det Y(t)$, then $\dot{d}(t) = \det(Y(t)) \operatorname{tr} (Y^{-1}(t) X Y(t)) = \operatorname{tr} X \det Y(t) = \operatorname{tr} X d(t)$.
Hence $d(1) = d(0) e^{\operatorname{tr} X}$.
