$E$ is a vector space with dimension $n$ and $f:E\to E$ is a linear map and for every $p=1,2,3,...$ we have $f^p:Λ^{p}(E) \to Λ^{p}(E)$ which is defined as below
$$(f^p(\alpha))(x_1,x_2,...,x_p)=\alpha(f(x_1),f(x_2),...,f(x_p))$$ and $f^0=I$ is Identity map
show that $$\det(I_E+f)=\sum_{p=0}^{n}\mathrm{tr}(f^p)$$
I have tried to write matrix of $f$ and $f^p$ with respect to a basis of $E$ and corresponding ones to $Λ^{p}(E)$ and it seems simple , but I think maybe it can be solved with some other way.
Let $\epsilon \in \bigwedge^n E$. Then for any $A: E \mapsto E$, the determinant $\det A$ obeys
$$A^n(\epsilon) = (\det A) \epsilon$$
Now do this for $A = I + f$, and let $X = x_1 \wedge x_2 \wedge \ldots \wedge x_n$:
$$\det(I + f) \epsilon(X) =\epsilon [ (I+f)(x_1) \wedge (I+f)(x_2) \wedge \ldots \wedge (I+f)(x_n)]$$
Now, suppose the vectors $x_i$ form an orthogonal set. We'll get terms like
$$\epsilon(X) + \epsilon[f(x_1) \wedge x_2 \ldots \wedge x_n] + \epsilon[x_1 \wedge f(x_2) \wedge \ldots \wedge x_n] + \ldots$$
Consider the second term. When the $x_i$ are orthonormal (normality is not required, but done for convenience here), it becomes clear that the only component of $f(x_1)$ that matters is the $x_1$ component of that vector; the other components are annihilated on the wedge. So this term becomes $\epsilon[(x_1) \cdot f(x_1) X]$. Carrying this out over all the first-order terms yields $\sum_{i} \epsilon[x_i \cdot f(x_i) X]$. That is clearly a trace; we're summing the diagonal components of the corresponding matrix.
All the other terms would turn out exactly the same way.
Incidentally, you also prove that $\det f = \mathrm{tr} f^n$ as a result.