Index Laws/Product of Indices/Semigroup

Theorem
Let $\left({T, \oplus}\right)$ be a semigroup.

For $a \in T$, let $\oplus^n a = a^n$ be defined as the $n$th power of $a$:


 * $a^n = \begin{cases}

a & : n = 1 \\ a^x \oplus a & : n = x + 1 \end{cases}$

That is:
 * $a^n = \underbrace{a \oplus a \oplus \cdots \oplus a}_{n \text{ copies of } a} = \oplus^n \left({a}\right)$

Then:
 * $\forall m, n \in \N_{>0}: a^{n m} = \left({a^n}\right)^m = \left({a^m}\right)^n$

That is:
 * $\forall m, n \in \N_{>0}: \oplus^{n m} a = \oplus^m \left({\oplus^n a}\right) = \oplus^n \left({\oplus^m a}\right)$

Proof
Let $b = \oplus^m a$.

Let $h: \N_{>0} \to T$ be the mapping defined as:


 * $\forall n \in \N_{>0}: h \left({n}\right) = \oplus^{n m} a$

Let the mapping $f_b: \N_{>0} \to T$ be recursively defined as:


 * $\forall n \in \N_{>0}: f_b \left({n}\right) = \oplus^n b$

From Uniqueness of Recursive Mapping from Naturally Ordered Semigroup to Algebraic Structure:
 * $f_b$ is the unique mapping which satisfies:
 * $\forall n \in \N_{>0}: f_b \left({n}\right) = \begin{cases}

b & : n = 1 \\ f_b \left ({r}\right) \oplus b & : n = r \circ 1 \end{cases}$

But $h \left({1}\right) = \oplus^{1 \times m} a = \oplus^m a = b$.

So:

Thus $h = f_b$, and so:


 * $\forall n, m \in \N_{>0}: \oplus^{n m} = \oplus^n \left({\oplus^m a}\right)$

From Natural Number Multiplication is Commutative:


 * $\forall n, m \in \N_{>0}: \oplus^m \left({\oplus^n a}\right) = \oplus^{m n} = \oplus^{n m}$