Exponentiation of natural numbers, $m^n$, and its agreement with the integer power in $\mathbb{R}$

Let $m\in \mathbb{N}$. By the recursion theorem (The recursion theorem) applied to the set $\mathbb{N}$, the starting element $1$ and the function $f(x)=x\cdot m$ (Multiplication of natural numbers), there is a unique function $\mathbb{N}\to \mathbb{N}$, written $n\mapsto m^n$, with

$m^0=1,m^{\sigma (n)}=m^n\cdot m(n\in \mathbb{N})\mathrm{.}$

Both the base and the value are natural numbers, so $m^n\in \mathbb{N}$ for all $m,n$. In particular $m^1=m^0\cdot m=m$ and $m^2=m\cdot m$.

Why a new item is needed. Integer powers $a^m$ defines $a^n$ for a real base $a$, so its value is a real number. The counts on this page, $\mid A^B\mid$ and $\mid \mathcal{P}(A)\mid$ among them, are natural numbers, and an identity between them has to be an identity in $\mathbb{N}$. The two operations are related by clause (d) below and by nothing weaker.

(a) $0^0=1$ and $0^n=0$ for $n\ge 1$. The first is the base clause. For the second, $0^{\sigma (n)}=0^n\cdot 0=0$, the clause $x\cdot 0=0$ being definitional (Multiplication of natural numbers), and every $n\ge 1$ is a successor.

(b) $1^n=1$ for every $n$. Induction: $1^0=1$, and $1^{\sigma (n)}=1^n\cdot 1=1^n$ (Zero and one under multiplication, The principle of mathematical induction).

(c) $m^{p+q}=m^p{m}^q$ and $(mp{)}^q=m^q{p}^q$. Both by induction on $q$, using associativity and commutativity of multiplication (Multiplication is associative, Multiplication is commutative). For the first, at $q=0$ we have $m^{p+0}=m^p=m^p\cdot 1=m^p{m}^0$, and $m^{p+\sigma (q)}=m^{\sigma (p+q)}=m^{p+q}\cdot m=(m^p{m}^q)\cdot m=m^p{m}^{\sigma (q)}$, using $p+\sigma (q)=\sigma (p+q)$ (Addition of natural numbers). For the second, at $q=0$ both sides are $1$, and $(mp{)}^{\sigma (q)}=(mp{)}^q(mp)=m^q{p}^{q}mp=m^{\sigma (q)}{p}^{\sigma (q)}$.

(d) The bridge into $\mathbb{R}$. With $\iota :\mathbb{N}\to \mathbb{R}$ the canonical natural (The canonical natural $\iota (n)=n\cdot 1_F$ of a field) and $x^n$ the integer power of Integer powers $a^m$,

$\iota \left(m^n\right)=\iota (m{)}^n(m,n\in \mathbb{N})\mathrm{.}$

Induction on $n$: at $n=0$ both sides are $1$, since $\iota (1)=1$; and $\iota (m^{\sigma (n)})=\iota (m^n\cdot m)=\iota (m^n)\iota (m)=\iota (m{)}^n\iota (m)=\iota (m{)}^{\sigma (n)}$, the second equality being the multiplicativity of $\iota$ (clause 0 of Laws of finite sums and products in $\mathbb{N}$, and $\iota \left({\sum }_{k<n}{a}_k\right)={\sum }_{k<n}\iota (a_k)$) and the last the recursion clause of Integer powers $a^m$.

(e) $m^n$ is a constant product. $m^n={\prod }_{k<n}m$, the $\mathbb{N}$-valued product of the constant list (Finite sums and finite products of natural numbers, ${\sum }_{k<n}{a}_k$ and ${\prod }_{k<n}{a}_k$ in $\mathbb{N}$). Induction: at $n=0$ both sides are $1$, and ${\prod }_{k<\sigma (n)}m=\left({\prod }_{k<n}m\right)\cdot m=m^n\cdot m=m^{\sigma (n)}$.

The recursion theorem

Multiplication of natural numbers

Integer powers $a^m$

Zero and one under multiplication

The principle of mathematical induction

Multiplication is associative

Multiplication is commutative

Addition of natural numbers

The canonical natural $\iota (n)=n\cdot 1_F$ of a field

Laws of finite sums and products in $\mathbb{N}$, and $\iota \left({\sum }_{k<n}{a}_k\right)={\sum }_{k<n}\iota (a_k)$

Finite sums and finite products of natural numbers, ${\sum }_{k<n}{a}_k$ and ${\prod }_{k<n}{a}_k$ in $\mathbb{N}$

• $0^0=1$, $0!=1$ and the empty product are one convention, not three. The value $0^0=1$ here is the base clause of the recursion above; by clause (e) it is the empty product of Finite sums and finite products of natural numbers, ${\sum }_{k<n}{a}_k$ and ${\prod }_{k<n}{a}_k$ in $\mathbb{N}$; and Integer powers $a^m$ adopts $a^0=1$ for every real $a$, $0$ included, so clause (d) is consistent at $m=n=0$. The reasons for the convention are set out in Integer powers $a^m$ and are not repeated here.

• The laws are the same laws. Clause (c) is the $\mathbb{N}$-valued form of clause 1 of Laws of integer exponents, which states $a^{m+n}=a^m{a}^n$, $(a^m{)}^n=a^{mn}$ and $(ab{)}^n=a^n{b}^n$ for a base in a field. Only the two identities actually used on this page are proved above; the third is available in $\mathbb{R}$ through clause (d) whenever it is wanted.

• The exponent stays a natural number. Following the convention of Finite sums and finite products, by recursion, the identification of a natural with its canonical natural is deliberately not made in an exponent: in $m^n$ and in $\iota (m{)}^n$ the exponent $n$ is a natural number, never a real.