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: $g{h}^0=ge=eg=h^{0}g$.

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 $k\mathrm{\ne }0$ 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 $c\mathrm{\ne }0$, in which case $x=-\iota (c)$ and $g^a(g^b{)}^{-1}=g^a(g^a{g}^c{)}^{-1}=g^a((g^c{)}^{-1}(g^a{)}^{-1})=((g^c{)}^{-1}{g}^a)(g^a{)}^{-1}=(g^c{)}^{-1}=g^x$, 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, $g^{x+y}=g^{a+c}(g^{b+d}{)}^{-1}=(g^a{g}^c)(g^b{g}^d{)}^{-1}=(g^a{g}^c)((g^d{)}^{-1}(g^b{)}^{-1})$. On the other side $g^x{g}^y=(g^a(g^b{)}^{-1}\left)\right(g^c(g^d{)}^{-1})=(g^a{g}^c)((g^b{)}^{-1}(g^d{)}^{-1})$, 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 $(gh{)}^n=((gh{)}^{-n}{)}^{-1}=(g^k{h}^k{)}^{-1}=(h^k{)}^{-1}(g^k{)}^{-1}=h^n{g}^n$ 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 $(g^x{)}^y=((g^x{)}^{-y}{)}^{-1}$, which by step 5.2 equals $(g^{x(-y)}{)}^{-1}=(g^{-(xy)}{)}^{-1}=((g^{xy}{)}^{-1}{)}^{-1}=g^{xy}$, 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.