T is expansive iff $T^n$ is expansive the definition of expansive transformation used is the following:
Let X be a compact metric space and $T: X → X$ a homeomorphism. T is said to be expansive if there exist a $δ > 0$ such that: if
x does not equal y, then there exists an $n ∈ Z$ such that $d(T^n
(x), T^n(y)) > δ.$ $δ$ is called an expansive constant for T.
Here is what I have so far: First for the case where T is expansive, 
I know that since T is a homemorphism then $T^n$ will be one too as composition of hemoemorphism is a homemorphism. Now I tried to use the same $\delta$. Fixing x and y not equal, I need to come up with an m s.t. Definition 7.1.4 Let X be a compact metric space and T : X −→ X a
homeomorphism. T is said to be expansive if there exist a δ > 0 such that: if
x not equal to y then there exists an n ∈ Z such that $d((T^n)^m(x), (T^n)^m(y))>\delta$ 
or $d(T^{nm}(x),T^{nm}(y))>\delta$. However that wont work for all m and n as sometimes nm wont be a multiple of the r required for $T^r(x)$ and $T^r(y)$ to be bigger than $\delta$ apart. Any hints?
 A: Let's prove the easy direction first: if $T^n$ is expansive, then $T$ is expansive. As $T^n$ is expansive, we have some $\delta > 0$ such that $x\neq y$ implies that there exists an $m\in \mathbb{Z}$ such that $$d((T^n)^m(x), (T^n)^m(y)) = d(T^{nm}(x), T^{nm}(y)) > \delta$$ Then, as $nm\in \mathbb{Z}$, we have some $n_1\in \mathbb{Z}$ such that $d(T^{n_1}(x), T^{n_1}(y)) > \delta$, and so $T$ is expansive. Now, we show that if $T$ is expansive, then $T^n$ is expansive. As $X$ is a compact metric space, $T$ is uniformly continuous. We define $\delta : \mathbb{R}^+\to \mathbb{R}^+$ such that $d(x, y)\leq \delta(\epsilon)$ implies $d(T(x), T(y))\leq \epsilon$. We choose $\delta_1$ as in the condition for $T$ to be expansive. Then, for $x\neq y$, we let $m\in \mathbb{Z}$ such that $d(T^m(x), T^m(y)) > \delta_1$. Consider that there is a $k\in \mathbb{Z}$ such that $0\leq m-kn\leq n-1$. We let $$\delta_n\leq \min\{\delta_1, \delta(\delta_1), \delta(\delta(\delta_1)), \ldots, \delta^{n-1}(\delta_1)\}$$ Then, as $$d(T^{m-kn}(T^{kn}(x)), T^{m-kn}(T^{kn}(y))) = d(T^m(x), T^m(y)) > \delta_1$$ we cannot have that $$d(T^{kn}(x), T^{kn}(y)) = d((T^n)^k(x), (T^n)^k(y))\leq \delta_n$$ as this would violate the uniform continuity of $T^{m-kn}$ (namely, $\delta_n\leq \delta^{m-kn}(\delta_1)$, so $d(T^{kn}(x), T^{kn}(y))\leq \delta_n$ would imply $d(T^m(x), T^m(y))\leq \delta_1$). Therefore, $\delta_n$ satisfies the condition for $T^n$ to be expansive.
