Laws of integer exponents

Base cases at $n = 0$ for the addition law, the product law and nonvanishing: $a^{m+0} = a^m = a^m \cdot 1 = a^m a^0$ for every $m \in \mathbb{N}$; $(ab)^0 = 1 = 1 \cdot 1 = a^0 b^0$; and if $a \ne 0$ then $a^0 = 1 \ne 0$.

Inductive hypothesis: fix $n \in \mathbb{N}$ and assume $a^{m+n} = a^m a^n$ for all $m \in \mathbb{N}$, $(ab)^n = a^n b^n$, and $a^n \ne 0$ whenever $a \ne 0$. The iterated-power law $(a^m)^n = a^{mn}$ is deliberately NOT carried in this hypothesis: its successor step needs the addition law at the exponent pair $(mn, m)$, whose second entry is not the current stage, so that law must be finished first and the iterated law proved afterwards.

For $a \ne 0$ and every integer $j$, $a^{-j} = (a^j)^{-1}$: for $j \ge 0$ this is the definition together with the agreement of the two clauses at $j = 0$, and for $j = -k$ with $k \ge 1$ it reads $a^{k} = (a^{-k})^{-1}$, which holds because $a^{-k} = (a^k)^{-1}$ and $(x^{-1})^{-1} = x$ at $x = a^{k}$. That last substitution needs $a^{k} \ne 0$, which is NOT free here and must not be read off the definition, since the definition of the negative clause is what is being justified; it is instead a self-contained induction on $k$, from $a^{0} = 1 \ne 0$ and the fact that $a^{k+1} = a^{k}a$ is a product of two nonzero elements of a field, hence nonzero.

Successor step for the addition law, the product law and nonvanishing: $a^{m+(n+1)} = a^{(m+n)+1} = a^{m+n} a = (a^m a^n) a = a^m (a^n a) = a^m a^{n+1}$ for every $m \in \mathbb{N}$; $(ab)^{n+1} = (ab)^n (ab) = (a^n b^n)(ab) = (a^n a)(b^n b) = a^{n+1} b^{n+1}$; and if $a \ne 0$ then $a^{n+1} = a^n a$ is a product of two nonzero elements, hence nonzero.

By the induction principle, for all $m, n \in \mathbb{N}$: $a^{m+n} = a^m a^n$ and $(ab)^n = a^n b^n$, and $a^n \ne 0$ whenever $a \ne 0$. The addition law is thereby available at EVERY pair of natural exponents, which is exactly what the iterated-power law needs.

The iterated-power law for natural exponents, $(a^m)^n = a^{mn}$, by a second induction on $n$ with $m \in \mathbb{N}$ fixed: at $n = 0$ both sides are $1$, since $(a^m)^0 = 1 = a^0 = a^{m \cdot 0}$; and if $(a^m)^n = a^{mn}$ then $(a^m)^{n+1} = (a^m)^n a^m = a^{mn} a^m = a^{mn+m} = a^{m(n+1)}$, where the third equality is the addition law of step 3.1 at the pair $(mn, m)$, legitimate precisely because that law is by now proved for all pairs of naturals. This completes claim 1.

For $a \ne 0$ and every integer $j$, $a^{j+1} = a^j a$: for $j \ge 0$ this is the recursion clause, and for $j = -k$ with $k \ge 1$ we compute $a^{-k} a = (a^k)^{-1} a = (a^{k-1} a)^{-1} a = (a^{k-1})^{-1} a^{-1} a = (a^{k-1})^{-1} = a^{-(k-1)} = a^{-k+1}$.

For $a, b \ne 0$ the product law holds for all integers $n$: for $n \ge 0$ it is step 3.1, and for $n = -k$ with $k \ge 1$ we get $(ab)^{-k} = ((ab)^k)^{-1} = (a^k b^k)^{-1} = (a^k)^{-1}(b^k)^{-1} = a^{-k} b^{-k}$.

For $a \ne 0$, every integer $j$ and every natural $k$, $a^{j+k} = a^j a^k$, by induction on $k$: the case $k = 0$ is $a^j = a^j \cdot 1$, and if $a^{j+k} = a^j a^k$ then $a^{j+k+1} = a^{j+k} a = a^j a^k a = a^j a^{k+1}$ by step 4.2 applied to the integer $j+k$ and by the recursion clause.

For $a \ne 0$ the addition law holds for all integers $m, n$: writing $n = k$ or $n = -k$ with $k \in \mathbb{N}$, the case $n = k$ is step 5.1, while for $n = -k$ step 5.1 applied to the integer $m - k$ gives $a^m = a^{(m-k)+k} = a^{m-k} a^k$, hence $a^{m-k} = a^m (a^k)^{-1} = a^m a^{-k}$.

For $a \ne 0$ the iterated-power law holds for all integers $m, n$: for $n \ge 0$ induction on $n$ gives $(a^m)^{n+1} = (a^m)^n a^m = a^{mn} a^m = a^{mn+m} = a^{m(n+1)}$, the third equality by the integer addition law of step 6.1 at the pair $(mn, m)$, with base $(a^m)^0 = 1 = a^0$; and for $n = -q$ with $q \ge 1$, $(a^m)^{-q} = ((a^m)^q)^{-1} = (a^{mq})^{-1} = a^{-mq} = a^{m(-q)}$, using that $a^m \ne 0$ by step 3.1 and step 1.3.

Claims 1, 2 and 3 are therefore established: the addition, product and iterated-power laws for natural exponents together with nonvanishing by steps 3.1 and 4.1, the identity $a^{-m} = (a^m)^{-1}$ by step 1.3, and the three integer-exponent laws by steps 6.1, 4.3 and 7.1.