Index Laws/Product of Indices/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:
 * $(a):\quad \forall a \in \F^* : \forall n, m \in \Z : \paren{a^m}^n = a^\paren{mn}$
 * $(b):\quad \forall a \in \F : \forall n, m \in \Z_{\ge 0} : \paren{a^m}^n = a^\paren{mn}$

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 Product of Powers of Group Elements:
 * $\forall a \in \F^* : \forall n, m \in \Z : \paren{a^m}^n = a^\paren{mn}$

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: $m \ne 0, n = 0$
Let $m \ne 0$ and $n = 0$.

Hence:
 * $m n = 0$

We have:

Case 3: $m \ne 0, n = 0$
Let $m \ne 0$ and $n \ne 0$.

Hence:
 * $m n \ne 0$

We have:

In all cases:
 * $\paren{\paren{0_F}^m}^n = \paren{0_F}^\paren{mn}$

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