I'm reading about the Thom form from in the paper: The Gauss-Bonnet-Chern Theorem on Riemannian Manifolds by Yin Li.
In page 31, he defines the Berezin integral on a real vector space $V$ to be a nonzero linear map $B: \wedge^*V \rightarrow \mathbb{R}$ which vanishes on $\wedge^kV$ for $k < dim_\mathbb{R}V$ and he gives a canonical example of it given by projecting $\omega \in \wedge^*V^* $ onto $e_1 \wedge ... \wedge e_n$ where $e_1,...,e_n$ is a basis of $V$. Then he says :
Since the identity map $Id : V \rightarrow V$ can be identified with an element of $\Gamma (V, \wedge^0 V^* \otimes V)$, we can take its exterior differential $dx \in \Gamma(V, \wedge^1 \otimes V)$. Then the exponential $e^{idx}$ lies in $\Gamma (V, \wedge^*V^* \otimes \wedge^*V^*.$ We extend the Berezin integral to a map $B: \Gamma(V, \wedge^*V^* \otimes \wedge^*V^* \rightarrow \Gamma(V, \wedge^*V^*)$ by $$ B(\omega \otimes \xi)= \omega \otimes B(\xi) , \omega \in \Gamma(V,\wedge^*V^*), \xi \in \wedge^*V^*. $$
My problem is that I didn't understand the following steps of the proof of proposition 4.3.2
Proof: Let $\lbrace e_k \rbrace$ be the dual basis of $dx^k,$ we have $$B(e^{-idx}) = B(\prod_{k=1}^n (1- idx^k \otimes e_k)) = {(-i)}^n B ((dx^1 \otimes e_1) \wedge...\wedge (dx^n \otimes e_n)).$$
My understanding of $dx$ which is viewed as an element of $\Gamma (V, \wedge^1 V^* \otimes V)$ is to be $dx = dx^1 \otimes e_1 +...+ dx^n \otimes e_n$ but I don't know how he calculates $e^{idx}$ to get the formulas above ?