Contraction Mapping Theorem
If $T\colon X\to X$ is a contraction mapping on a complete metric space $(X,d)$ then there is exactly one solution $x\in X$.
Proof:
Let $x_0$ be any point in $X$. We define a sequence by $$x_{n+1}=Tx_n, \qquad \text{for } n\geq 0.$$ Denote the $n$th iterate of $T$ by $T^n$, so that $x_n = T^n x_0$. First, we show that $(x_n)$ is a Cauchy Sequence. If $n\geq m\geq 1$ then $$d(x_n,x_m) = d(T^n x_0, T^m x_0) \leq c^md(T^{n-m} x_0,x_0).$$
This is only some of the proof but my question is about the last inequality in particular the $T^{n-m}$. How is this obtained?
Thank you for any help and comments.