Proof that the determinant of a matrix exponential equals the exponential of trace of the matrix The question arises from this tweet:

 A: It was supposed to be a question, but I kept going...
I believe that the idea would be to use the diagonalization of the square matrix $M,$ such as
$$M= P\Lambda P^{-1}$$
where $$\Lambda =\begin{bmatrix}\lambda_1&0&\cdots&0\\0&\lambda_2&\cdots&0\\\vdots&\cdots&\ddots&\vdots\\0&\cdots&\cdots&\lambda_n\end{bmatrix}$$
is the matrix of eigenvalues.
Now since
$$\begin{align}
e^M &= \sum_{n=0}^\infty \frac{M^n}{n!}\\
&= \sum_{n=0}^\infty \frac{P\Lambda^n P^{-1}}{n!}\\
&=  P\left(\sum_{n=0}^\infty \frac{\Lambda^n }{n!}\right)P^{-1}\\
&=Pe^\Lambda P^{-1}
\end{align}$$
The determinant will be
$$\begin{align}
\det(e^M) & =\det( P) \det \left( e^\Lambda \right )\det \left( P^{-1}\right)\\
&= \det \left( e^\Lambda \right )\\
&= e^{\lambda_1} e^{\lambda_2}\cdots e^{\lambda_n}\\
&=e^{\lambda_1 + \lambda_2+\cdots +\lambda_n}
\end{align}$$
The right side of the equation would be
$$\begin{align}
e^{\text{Tr} (M)} & = e^{\text{Tr} \left({P\Lambda P^{-1}}\right)}\\
&=e^{\text{Tr} {(\Lambda)} }\\
&=e^{\lambda_1 + \lambda_2+\cdots +\lambda_n}
\end{align}
$$
