Let $p$ be a periodic point of period $m$ for a $C^1$ map $f$. Suppose that the differential $Df^m_p$ does not have 1 as an eigenvalue. Let us introduce local coordinates near $p$ with $p$ as the origin. In this coordinates $Df^m_0$ becomes a matrix. Consider the following fact:
Since $1$ is not among the eigenvalues of $Df^m_0$ the map $F = f^m - Id$ defined locally in these coordinates is locally invertible by the Inverse Function Theorem.
Can someone explain me why does the fact that $1$ is not an eigenvalue imply that $F$ is invertible?
If $D f _p^m$ does not have $1$ as an eigenvalue, then $DF = Df _p ^m - I$ (where $I$ denotes the identity matrix) does not have $0$ as an eigenvalue, making $DF$ an isomorphism between $T_p M$ and $T_{F(p)}M$.
Thus $F$ is locally invertible by the inverse function theorem (see for example https://en.wikipedia.org/wiki/Inverse_function_theorem#Manifolds).