Repeated Composition of Injection is Injection

Theorem
Let $S$ be a set.

Let $f: S \to S$ be an injection.

Let the sequence of mappings:
 * $f^0, f^1, f^2, \ldots, f^n, \ldots$

be defined as:
 * $\forall n \in \N: f^n \left({x}\right) = \begin{cases}

x & : n = 0 \\ f \left({x}\right) & : n = 1 \\ f \left({f^{n-1} \left({x}\right)}\right) & : n > 1 \end{cases}$

Then for all $n \in \N$, $f^n$ is an injection.

Proof
Proof by induction:

For all $n \in \N_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $f^n$ is an injection.

$P \left({0}\right)$ is true, as this is the case Identity Mapping is Injection.

$P \left({1}\right)$ is true, as this is the assertion that $f$ is an injection.

Basis for the Induction
$P \left({2}\right)$ is the assertion that $f^2$ is an injection.

But $f \circ f$ is an injection by Composite of Injections is Injection.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $f^k$ is an injection.

Then we need to show:
 * $f^{k+1}$ is an injection.

Induction Step
This is our induction step:

By the induction hypothesis $f^k$ is an injection.

By Composite of Injections is Injection it follows that $f^{k+1}$ is an injection.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore for all $n \in \N$:
 * $f^n$ is an injection.