Exponent laws in a group: $g^{m+n} = g^{m}g^{n}$ and $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$, and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute

Natural exponents, base of claim 1: for every $m \in \mathbb{N}$, $g^{m+0} = g^{m} = g^{m} e = g^{m} g^{0}$.

Natural exponents, inductive hypothesis for claim 1: fix $n \in \mathbb{N}$ and assume $g^{m+n} = g^{m} g^{n}$ for every $m \in \mathbb{N}$ and every $g$.

Suppose $gh = hg$; base of the commuting sub-claim: $g h^{0} = g e = e g = h^{0} g$.

Inductive hypothesis of the commuting sub-claim: assume $g h^{n} = h^{n} g$.

Base of claim 5 for natural exponents: $(gh)^{0} = e = e \cdot e = g^{0} h^{0}$.

Inductive hypothesis of claim 5 for natural exponents: assume $gh = hg$ and $(gh)^{n} = g^{n} h^{n}$.

If $uv = vu$ in $G$ then $u v^{-1} = v^{-1} u$: multiplying $uv = vu$ on the left and on the right by $v^{-1}$ gives $v^{-1}(uv)v^{-1} = v^{-1}(vu)v^{-1}$, and regrouping both sides gives $v^{-1} u = u v^{-1}$. Consequently $u^{-1} v^{-1} = (vu)^{-1} = (uv)^{-1} = v^{-1} u^{-1}$.

Claim 2 for a nonnegative exponent. If $k = 0$ then $-\iota(0) = 0$, so $g^{-\iota(0)} = g^{0} = e$ and $(g^{0})^{-1} = e^{-1} = e$. If $k \ne 0$ then $0 < \iota(k)$ because $\iota$ preserves the order, so $-\iota(k) < 0$ and the second clause of the definition gives $g^{-\iota(k)} = (g^{k})^{-1}$ directly.

Claim 3 for a nonnegative second exponent, base: $(g^{x})^{0} = e = g^{0} = g^{x \cdot 0}$, since $x \cdot 0 = 0$ in $\mathbb{Z}$.

Claim 3 for a nonnegative second exponent, inductive hypothesis: assume $(g^{x})^{\iota(n)} = g^{x \iota(n)}$ for this $n \in \mathbb{N}$ and every $x \in \mathbb{Z}$.

Successor step for claim 1 with natural exponents: $g^{m+\sigma(n)} = g^{\sigma(m+n)} = g^{m+n} g = (g^{m} g^{n}) g = g^{m}(g^{n} g) = g^{m} g^{\sigma(n)}$; by induction, $g^{m+n} = g^{m} g^{n}$ for all $m, n \in \mathbb{N}$, in any monoid.

Successor step for the commuting sub-claim: $g h^{\sigma(n)} = g(h^{n} h) = (g h^{n})h = (h^{n} g)h = h^{n}(gh) = h^{n}(hg) = (h^{n} h)g = h^{\sigma(n)} g$; by induction $g h^{n} = h^{n} g$ for every $n \in \mathbb{N}$ whenever $gh = hg$.

Claim 2 in general. If $0 \le x$ then $x = \iota(k)$ and step 1.8 gives $g^{-x} = (g^{k})^{-1} = (g^{x})^{-1}$. If $x < 0$ then $0 < -x$, so $-x = \iota(k)$ and by definition $g^{x} = (g^{k})^{-1}$, whence $(g^{x})^{-1} = g^{k} = g^{-x}$. This is claim 2.

Successor step for claim 5 with natural exponents: assuming $gh = hg$, $(gh)^{\sigma(n)} = (gh)^{n}(gh) = (g^{n} h^{n})(gh) = g^{n}((h^{n} g)h) = g^{n}((g h^{n})h) = (g^{n} g)(h^{n} h) = g^{\sigma(n)} h^{\sigma(n)}$, the fourth equality being step 2.2 applied to $g$ and $h^{n}$; by induction $(gh)^{n} = g^{n} h^{n}$ for every $n \in \mathbb{N}$.

Normal form. Let $x \in \mathbb{Z}$ and write $x = \iota(a) - \iota(b)$ with $a, b \in \mathbb{N}$, possible since $x = [(a,b)]$ for some naturals. Then $g^{x} = g^{a}(g^{b})^{-1}$. Indeed, by trichotomy either $b \le a$, say $a = b + c$, in which case $x = \iota(c)$ and $g^{a}(g^{b})^{-1} = (g^{b} g^{c})(g^{b})^{-1} = (g^{c} g^{b})(g^{b})^{-1} = g^{c} = g^{x}$, using step 2.1 twice together with commutativity of addition on $\mathbb{N}$; or $a < b$, say $b = a + c$ with $c \ne 0$, in which case $x = -\iota(c)$ and $g^{a}(g^{b})^{-1} = g^{a}(g^{a} g^{c})^{-1} = g^{a}\bigl((g^{c})^{-1}(g^{a})^{-1}\bigr) = \bigl((g^{c})^{-1} g^{a}\bigr)(g^{a})^{-1} = (g^{c})^{-1} = g^{x}$, the third equality using step 1.7 with $u = g^{a}$ and $v = g^{c}$, which commute by step 2.1.

Claim 1 for integer exponents. Write $x = \iota(a) - \iota(b)$ and $y = \iota(c) - \iota(d)$, so $x + y = \iota(a+c) - \iota(b+d)$. By step 3.2 and step 2.1, $g^{x+y} = g^{a+c}(g^{b+d})^{-1} = (g^{a} g^{c})(g^{b} g^{d})^{-1} = (g^{a} g^{c})\bigl((g^{d})^{-1}(g^{b})^{-1}\bigr)$. On the other side $g^{x} g^{y} = \bigl(g^{a}(g^{b})^{-1}\bigr)\bigl(g^{c}(g^{d})^{-1}\bigr) = (g^{a} g^{c})\bigl((g^{b})^{-1}(g^{d})^{-1}\bigr)$, moving $(g^{b})^{-1}$ past $g^{c}$ by step 1.7, since $g^{b}$ and $g^{c}$ commute by step 2.1. Finally $(g^{b})^{-1}(g^{d})^{-1} = (g^{d})^{-1}(g^{b})^{-1}$ by step 1.7 again, so the two sides agree.

Claim 5 for a negative exponent: assume $gh = hg$ and let $n < 0$, so $-n = \iota(k)$ with $k \in \mathbb{N}$. Then $(gh)^{n} = \bigl((gh)^{-n}\bigr)^{-1} = (g^{k} h^{k})^{-1} = (h^{k})^{-1}(g^{k})^{-1} = h^{n} g^{n}$ by claim 2 and step 3.1; and $g^{k}$ and $h^{k}$ commute, by step 2.2 applied twice, so their inverses commute by step 1.7, giving $h^{n} g^{n} = g^{n} h^{n}$. With step 3.1 this proves claim 5 for every $n \in \mathbb{Z}$.

Claim 4. By claim 1 and commutativity of addition in $\mathbb{Z}$, $g^{x} g^{y} = g^{x+y} = g^{y+x} = g^{y} g^{x}$.

Claim 3, successor step: $(g^{x})^{\sigma(n)} = (g^{x})^{n} g^{x} = g^{x\iota(n)} g^{x} = g^{x\iota(n) + x} = g^{x(\iota(n) + 1)} = g^{x \iota(\sigma(n))}$, using the hypothesis, then claim 1, then distributivity in $\mathbb{Z}$, then $\iota(\sigma(n)) = \iota(n+1) = \iota(n) + 1$. By induction, $(g^{x})^{y} = g^{xy}$ for every $x \in \mathbb{Z}$ and every $y \ge 0$. When $x = \iota(m)$ is itself nonnegative the two exponents $x\iota(n)$ and $x$ occurring here are nonnegative as well, so the appeal to claim 1 is an appeal to its monoid form, step 2.1, and the computation uses no inverse; that is the natural-exponent case, valid in any monoid.

Claim 3 for a negative second exponent: let $y < 0$, so $0 < -y$. Applying claim 2 to the element $g^{x}$ gives $(g^{x})^{y} = \bigl((g^{x})^{-y}\bigr)^{-1}$, which by step 5.2 equals $\bigl(g^{x(-y)}\bigr)^{-1} = \bigl(g^{-(xy)}\bigr)^{-1} = \bigl((g^{xy})^{-1}\bigr)^{-1} = g^{xy}$, using claim 2 once more and then $(u^{-1})^{-1} = u$. Together with step 5.2 this is claim 3.

Claims 1 to 5 are established: claim 1 in step 4.1, claim 2 in step 2.3, claim 3 in step 6.1, claim 4 in step 5.1 and claim 5 in steps 3.1 and 4.2; the natural-exponent forms of claims 1, 3 and 5 are steps 2.1, 5.2 and 3.1, and use no inverses.