Index Laws/Common Index/Field

Theorem
Let $\struct {F, +, \circ}$ be a field with zero $0_F$ and unity $1_F$.

Let $F^* = F \setminus {0_F}$ denote the set of elements of $F$ without the zero $0_F$.

Then:
 * $(a):\quad \forall a, b \in \F^* : \forall n \in \Z : a^n \circ b^n = \paren{ab}^n$
 * $(b):\quad \forall a, b \in \F : \forall n \in \Z_{\ge 0} : a^n \circ b^n = \paren{ab}^n$

Statement $(a)$
By :
 * $\struct{F^*, \circ}$ is an Abelian group

By :
 * For all $a \in F^*$ and $n \in \Z$, $a^n$ is defined as the $n$th power of $a$ with respect to the Abelian group $\struct {F^*, \circ}$

From Sum of Powers of Group Elements:
 * $\forall a \in \F^* : \forall n, m \in \Z : a^m \circ a^n = a^\paren{m + n}$

Statement $(b)$
Let $m,n \in \Z_{\ge 0}$ be arbitrary elements of $\Z_{\ge 0}$.

For $a \in F^*$, $(b)$ follows from $(a)$.

It remains to show that $(b)$ holds for $0_F$.

Case 1: $m = 0$
Let $m = 0$.

We have:

Case 2: $m \ne 0$
Let $m \ne 0$.

Hence:
 * $m + n \ne 0$

We have:

In both cases:
 * $\paren{0_F}^m \circ \paren{0_F}^n = \paren{0_F}^\paren{m + n}$

Since $m,n$ were arbitrary elements of $\Z_{\ge 0}$:
 * $\forall n, m \in \Z_{\ge 0} : \paren{0_F}^m \circ \paren{0_F}^n = \paren{0_F}^\paren{m + n}$