Let $\Lambda^p (E)$ be the set of $p$-covariant exterior tensors on linear space $E$ over field $K$ (dim$E=n$ and $ 0\leq p\leq n $ , $\Lambda^0E:=K$). We define linear map $f^p:\Lambda^p(E)\rightarrow \Lambda^p(E)$ induced form linear transformation $f:E\rightarrow E$ with the following rule:
$(f^p)(\alpha)(u_1,u_2,\cdots,u_p)=\alpha(f(u_1),\cdots,f(u_p))\qquad ,\forall\alpha\in \Lambda^p(E) ,$
and suppose that $f^0:K\rightarrow K$ is identical map.
Note that $I_E$ is identical map $I_E:E\rightarrow E$. Now how can I show
$\text{det}(I_E+f)=\displaystyle\sum_{p=0}^n\text{trace}(f^p)$ ?
I have tried to write matrices with respect to a basis of $E$ and corresponding ones to $\Lambda^p(E)$ and it seems simple , but I think maybe it can be solved with the following formula: $\text{det}(f)=\displaystyle\sum_{\sigma\in S_n}\text{sign}(\sigma) \ a_{1,\sigma(1)},\cdots a_{n,\sigma(n)}$ . But it seems that it is hard for working with this formula.