Ordinal exponentiation $\alpha^{\beta}$, with the conventions $\alpha^{0} = 1$ and $0^{0} = 1$

Let $\alpha$ and $\beta$ be ordinals (Ordinal (von Neumann)). The power $\alpha^{\beta}$ is defined by recursion on $\beta$, in the three cases of Successor and limit ordinals:

$\alpha^{0} = 1, \qquad \alpha^{\beta^{+}} = \alpha^{\beta} \cdot \alpha, \qquad \alpha^{\lambda} = \bigcup\{\, \alpha^{\beta} : 0 < \beta < \lambda \,\} \quad (\lambda \text{ a limit ordinal}),$

with $\cdot$ the ordinal multiplication of Ordinal multiplication $\alpha \cdot \beta$. That exactly one operation satisfies these three clauses, and that all its values are ordinals, is Ordinal exponentiation exists and is unique, with the limit clause taken over $0 < \beta < \lambda$ so that $0^{\lambda} = 0$, proved immediately above.

The first clause applies to every $\alpha$, so in particular $0^{0} = 1$.

• The limit clause ranges over $0 < \beta < \lambda$, not over $\beta < \lambda$, and the restriction is load bearing. The unrestricted union would include the value $\alpha^{0} = 1$, and at $\alpha = 0$ that single stray term flips the answer: it would make $0^{\omega} = 1$ instead of $0$. With the restriction, one formula is correct for every $\alpha$ at once and no case split on $\alpha$ is needed. Ordinal exponentiation exists and is unique, with the limit clause taken over $0 < \beta < \lambda$ so that $0^{\lambda} = 0$ carries the details of that restriction; $\alpha^{\beta+\gamma} = \alpha^{\beta}\cdot\alpha^{\gamma}$ and $(\alpha^{\beta})^{\gamma} = \alpha^{\beta\cdot\gamma}$; and for $\alpha > 1$ exponentiation is strictly increasing with $\beta \le \alpha^{\beta}$ proves the exponent law that the unrestricted clause would falsify. For $\alpha \ge 1$ the two clauses agree, because then $\alpha^{0} = 1 \le \alpha = \alpha^{1}$ and $1 < \lambda$, so dropping the term at $\beta = 0$ does not lower the supremum.

• $\alpha^{1} = \alpha$. Indeed $\alpha^{1} = \alpha^{0^{+}} = \alpha^{0} \cdot \alpha = 1 \cdot \alpha$, and $1 \cdot \alpha = \alpha$ is proved in Monotonicity of ordinal $+$ and $\cdot$: strictly increasing and continuous in the right argument, weakly increasing in the left, with left cancellation, and the identities $0 + \beta = \beta$ and $1 \cdot \beta = \beta$.

• This is not cardinal exponentiation. The ordinal $2^{\omega}$ is $\omega$, computed in FALSE: the ordinal $2^{\omega}$ is uncountable; the cardinal power of $2$ by the size of $\mathbb{N}$ is the size of $\mathcal{P}(\mathbb{N})$, which is uncountable. The two operations share one notation and are not the same function. Ordinal $\alpha^{\beta}$ and cardinal $\kappa^{\lambda}$ are different operations that share one notation is the standing warning, and cardinal exponentiation is not defined at this point in the reading order; it is introduced later, on Cardinal Arithmetic, Cofinality and the Alephs.

• Notation and precedence. $\alpha^{\beta} \cdot \gamma$ means $(\alpha^{\beta}) \cdot \gamma$, and $\alpha^{\beta} + \gamma$ means $(\alpha^{\beta}) + \gamma$; powers bind tightest, then products, then sums. The Cantor normal form of Cantor normal form: every nonzero ordinal is $\omega^{\beta_0}\cdot c_0 + \cdots + \omega^{\beta_{k-1}}\cdot c_{k-1}$ with $\beta_0 > \cdots > \beta_{k-1}$ and each $c_i$ a nonzero natural number, in exactly one way is written with that convention throughout.

• The base is a parameter, the exponent is what the recursion runs on. As with Ordinal addition $\alpha + \beta$ and Ordinal multiplication $\alpha \cdot \beta$, the recursion is on the right argument, and the operation is correspondingly asymmetric: $\alpha \mapsto \alpha^{\beta}$ and $\beta \mapsto \alpha^{\beta}$ behave quite differently, and only the second is continuous at limits.