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

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

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, $\lvert A^{B}\rvert$ and $\lvert\mathcal{P}(A)\rvert$ 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 $(m p)^{q} = m^{q} p^{q}$. 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 $a^m$,

$\iota\big(m^{n}\big) = \iota(m)^{n} \qquad (m, n \in \mathbb{N}).$

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 $\iota\big(\sum_{k<n} a_k\big) = \sum_{k<n} \iota(a_k)$) 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 $\prod_{k<\sigma(n)} m = \big(\prod_{k<n} m\big)\cdot m = m^{n}\cdot m = m^{\sigma(n)}$.

• $0^{0} = 1$, $0! = 1$ 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.