Prove that there is no $(x,n,m)$ such that $f^n(x)=f^{n+km}(x)$ if $f$ is injective Assume that $f : X\mapsto  X$ is injective. And, we write $f\circ f\circ\dots\circ f(x)=f^n(x)$ for $n$ times composition. Prove that there is no $x\in X$ and no $n,m\in \mathbb{N}$ such that the following equality holds:
$f^n(x)=f^{n+m}(x)=f^{n+2m}(x)=\dots$
It is an impenetrable problem for me, even don't know how to start. Of cousre, it's easy to show that if $f : X\mapsto  X$ is injective, then $f^n : X\mapsto X$ is injective for each $n\in \mathbb{N}$; but does it helps at all?
Thank you in advance.

EDIT: The original question:

Assume that $f : X \to X$ is injective.
a. Prove that $f^n : X \to X$ is injective for each $n \in \mathbb{Z}_+$.
b. Prove that $f$ has no eventual periodic points.

From the book: C. Adams, R. Franzosa, Introduction to Topology: Pure and Applied.
Also from the book: We say that $x$ is an eventual periodic point if $x$ is not a periodic
point but $f^n(x)$ is a periodic point for some $n\in \mathbb{Z_+}$.
 A: This is false. Take $f\colon x\mapsto x$ (identity function). For any $x\in X$, $n,m \in\mathbb{N}$ you have $f^{n}(x) =  x = f^{n+m}(x) = f^{n+km}(x)$ for all $k \geq 0$.
But $f$ is injective.
You may want another condition on $f$, e.g. "$f$ has no fixed point" (Edit: and even this is not enough, as pointed out in the comments below).
A: Suppose that $f^n(x)=f^{n+m}(x)$; since $f$ is injective you may 'cancel' to obtain $f^{n-1}(x)=f^{n-1+m}(x)$. Repeating this procedure eventually leads you to $x=f^m(x)$ which is a contradiction.
A: You made a mistake in interpreting the question. You want to show that $f$ has no eventual periodic points, that is, for all $x \in X$, $x$ is not an eventual periodic point. It's thus sufficient to prove that, for all $x$, if $f^n(x)$ is periodic for some $n > 1$, then $x$ itself is periodic. This isn't what you wrote: it's possible for some $f^n(x)$ to be periodic.
Now it's a bit easier to prove the correct statement. Suppose that $f^n(x)$ is periodic for some $n > 1$. Then there is an integer $m > 0$ such that $f^n(x) = f^{n+m}(x) = f^{n+2m}(x) = \dots$. But the map $f$ is injective, so $f^n(x) = f^{n+m}(x) = f^n(f^m(x)) \implies x = f^m(x)$. By induction, you actually find that $x = f^m(x) = f^{2m}(x) = \dots$, and thus $x$ is periodic. Thus, by your definition, $x$ is not eventual periodic.
