Homomorphism of Powers/Naturally Ordered Semigroup

Theorem
Let $\left({T_1, \odot_1}\right)$ and $\left({T_2, \odot_2}\right)$ be semigroups.

Let $\phi: \left({T_1, \odot_1}\right) \to \left({T_2, \odot_2}\right)$ be a (semigroup) homomorphism.

Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup.

Let $\odot_1^n$ and $\odot_2^n$ be as defined as in Recursive Mapping to Semigroup.

Then:
 * $\forall a \in T_1: \forall n \in \left({S^*, \circ, \preceq}\right): \phi \left({\odot_1^n \left({a}\right)}\right) = \odot_2^n \left({\phi \left({a}\right)}\right)$

Proof
Proof by the Principle of Finite Induction:

For all $n \in \left({S^*, \circ, \preceq}\right)$, let $P \left({n}\right)$ be the proposition:
 * $\phi \left({\odot_1^n \left({a}\right)}\right) = \odot_2^n \left({\phi \left({a}\right)}\right)$

Basis for the Induction
$P(1)$ is true, as this just says:

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $\phi \left({\odot_1^k \left({a}\right)}\right) = \odot_2^k \left({\phi \left({a}\right)}\right)$

Then we need to show:
 * $\phi \left({\odot_1^{k+1} \left({a}\right)}\right) = \odot_2^{k+1} \left({\phi \left({a}\right)}\right)$

Induction Step
This is our induction step:

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

Therefore, by the Principle of Finite Induction:
 * $\forall n \in \left({S^*, \circ, \preceq}\right): \phi \left({\odot_1^n \left({a}\right)}\right) = \odot_2^n \left({\phi \left({a}\right)}\right)$

as we wanted to show.