Given 1 is not an eigenvalue of $df$ at $x_0$, take a chart $(U,\phi)$ around $x_0.$ Then in this coordinate neighborhood, think of $f$ as a map from open ball in $\mathbb{R}^n$ (say $B$), to itself with $f(0)=0.$ Now consider we have a function $f:B\rightarrow B$ such that $f(0)=0.$ Then $det(f - id)(0)\neq 0$.
But I thought it should be det$(df-id)(0) \neq 0$? Did we assume f is linear map so f = df here?
This question is answered by Ted Shifrin.
Yes this should be det$(df - id)_0 \neq 0$. I can't assume that $f$ is linear.