Let $\displaystyle V$ be a m dimensional vector space over real numbers.
$\displaystyle A$ be a linear operator on $\displaystyle V$,
$\displaystyle \wedge^{p}A$ be a linear operator on p-th exterior algebra $\displaystyle \wedge^{p}V$,as defined by
$\displaystyle (\wedge^{p}A)(v_{1}\wedge...\wedge v_{p})=(Av_{1})\wedge...\wedge(Av_{p})$.
$\displaystyle D^{p}A$ is another linear operator, as defined by
$\displaystyle (D^{p}A)(v_{1}\wedge...\wedge v_{p})=\sum_{r=1}^{p}v_{1}\wedge...\wedge v_{r-1}\wedge(Av_{r})\wedge v_{r+1}\wedge...\wedge v_{p}$.
$\displaystyle v_{1},...,v_{p}\in V$.
We difine $\displaystyle \wedge^{0}A$ be the identity endomorphism,
$\displaystyle D^{0}A$ be the zero endomorphism of the scalars.

how can we get the formula?
$\displaystyle det(I-e^{xA})=\sum_{p=0}^{m}(-1)^{p}Tr(\wedge^{p}( e^{xA}))=\sum_{p=0}^{m}(-1)^{p}Tr(e^{xD^{p}A})$