Let $M$ be a compact $n-$ dimensional $\mathbb{R}$-manifold and $f \in C^{\infty}(M,M)$. Then $f$ induces the differential $Tf: TM \to TM$ on tangent spaces and therefore via pullback as well on dual spaces $Tf^*: T^*M \to T^*M$ which extends naturally to a map on exterior algebra $\wedge^{\bullet} T^*M$. spelling it out explicitely we have for every $x \in M$ a pullback map
$$f_x^*: \wedge^{\bullet} T^*_{f(x)} M \to \wedge^{\bullet} T^*_{x} M, \omega \mapsto f^* \omega$$
Assume $x \in M$ is fixed by $f$: that is $f(x)=x$. The exterior algebra $\wedge^{\bullet} T^*_{x}M = \bigoplus_q \wedge^{q} T^*_{x}M$ is a vector space and then $f_x^*$ acts as an endomorphism of this space, and we can build a trace of $f_x^*$.
My question is why this identity holds:
$$\sum_q ^n (-1)^q \cdot Tr \ (f_x^* \vert _{\wedge^{q} T^*_{x}M})= \det(1- T_x f)$$
As Qiaochu Yuan remarked this is purely a statement about vector spaces. That is let $V= \mathbb{R}^n$ a real vector space and $A \in End(V)$. Let $A^* \in End(V^*)$ the induced endomorphism of the dual space and call $\overline{A}$ the extension of $A^*$ to $\wedge^{\bullet} V^*$.
That is we have to show
$$\sum_q ^n (-1)^q \cdot Tr \ (\overline{A} \vert _{\wedge^{q} V^*})= \det(1- A)$$
The story looks nice if $A$ is diagonalizable. What, if $A$ is not diagonalizable. assume firstly we can diagonalize $A$ and $\lambda_i$ are eigenvalues. Then $\det(1- A)=\det(1-A)= \prod_i (1-\lambda_i)$.
We observe that the dual $A^*$ has same eigenvalues as $A$ and $Tr \ (\overline{A} \vert _{\wedge^{q} V^*})= \sum_i S_i^q(\lambda_1,..., \lambda_n)$ where $S_i^q$ are the $q$-th symmetric polynomials in $n$-variables. This imply the claim.
But what if $A$ isn't diagonalizable?