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 $a\mathrm{\ne }0$ then $a^0=1\mathrm{\ne }0$.

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 $a^n\mathrm{\ne }0$ whenever $a\mathrm{\ne }0$. 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 $a\mathrm{\ne }0$ 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 $a^k\mathrm{\ne }0$, 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 $a^0=1\mathrm{\ne }0$ 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 $a\mathrm{\ne }0$ 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 $a^n\mathrm{\ne }0$ whenever $a\mathrm{\ne }0$. 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 $a\mathrm{\ne }0$ 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 $a,b\mathrm{\ne }0$ 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 $a\mathrm{\ne }0$, 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 $a\mathrm{\ne }0$ 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 $a\mathrm{\ne }0$ 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 $a^m\mathrm{\ne }0$ 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.