Ordinal exponentiation exists and is unique, with the limit clause taken over $0<\beta <\lambda$ so that $0^{\lambda }=0$

Fix an ordinal $\alpha$ (Ordinal (von Neumann)). There is exactly one class function $\beta \mapsto {\mathrm{e}}_{\alpha }(\beta )$, defined at every ordinal $\beta$, satisfying the three clauses

${\mathrm{e}}_{\alpha }(0)=1,{\mathrm{e}}_{\alpha }({\beta }^{+})={\mathrm{e}}_{\alpha }(\beta )\cdot \alpha ,{\mathrm{e}}_{\alpha }(\lambda )=\bigcup \{{\mathrm{e}}_{\alpha }(\beta ):\beta \in \lambda \text{ and }\beta \mathrm{\ne }0\}(\lambda \text{ a limit ordinal}),$

with $\cdot$ the ordinal multiplication of Ordinal multiplication $\alpha \cdot \beta$, and every value ${\mathrm{e}}_{\alpha }(\beta )$ is an ordinal.

The limit clause runs over $0<\beta <\lambda$, and that restriction is not cosmetic. With the unrestricted clause ${\mathrm{e}}_{\alpha }(\lambda )=\bigcup \{{\mathrm{e}}_{\alpha }(\beta ):\beta \in \lambda \}$ the value ${\mathrm{e}}_0(0)=1$ would be one of the sets united, so ${\mathrm{e}}_0(\omega )$ would come out $\ge 1$ and in fact equal to $1$, whereas $0$ raised to a limit must be $0$. With the restriction above the single formula is correct for every $\alpha$, including $\alpha =0$, and no case split on $\alpha$ is needed. For $\alpha \ge 1$ the restriction changes nothing, since then ${\mathrm{e}}_{\alpha }(0)=1\le \alpha ={\mathrm{e}}_{\alpha }(1)$ and $1\in \lambda$.

This is the well-definedness obligation discharged before ordinal exponentiation is written down; the operation itself is named in the definition that follows. The proof is a theorem of ZF and uses no choice principle.

The value at $\alpha =0$, worked out, and what the naive clause breaks. By the clauses, ${\mathrm{e}}_0(0)=1$ and ${\mathrm{e}}_0(1)={\mathrm{e}}_0(0)\cdot 0=1\cdot 0=0$, and then ${\mathrm{e}}_0(\beta )=0$ for every $\beta >0$; at a limit $\lambda$ the restricted union is $\bigcup \{0\}=0$, as it should be. Had the union run over all $\beta \in \lambda$ it would have contained ${\mathrm{e}}_0(0)=1$, giving ${\mathrm{e}}_0(\omega )=\bigcup \{1,0\}=1$. That is not merely unattractive: it falsifies the exponent law ${\alpha }^{\beta +\gamma }={\alpha }^{\beta }\cdot {\alpha }^{\gamma }$ of ${\alpha }^{\beta +\gamma }={\alpha }^{\beta }\cdot {\alpha }^{\gamma }$ and $({\alpha }^{\beta }{)}^{\gamma }={\alpha }^{\beta \cdot \gamma }$; and for $\alpha >1$ exponentiation is strictly increasing with $\beta \le {\alpha }^{\beta }$ at $\alpha =0$, $\beta =1$, $\gamma =\omega$, since $1+\omega =\omega$ makes the left side $0^{\omega }=1$ while the right side is $0^1\cdot 0^{\omega }=0\cdot 1=0$. Many texts avoid the issue by splitting the definition into a case $\alpha =0$ and a case $\alpha >0$; the restricted clause is the same definition without the split.

The convention $0^0=1$. The clause ${\mathrm{e}}_{\alpha }(0)=1$ applies to every $\alpha$, so $0^0=1$ here. This is the convention that makes the successor clause uniform, and it is the one used in Ordinal exponentiation ${\alpha }^{\beta }$, with the conventions ${\alpha }^0=1$ and $0^0=1$ and everywhere below.

This is ordinal, not cardinal, exponentiation. The two operations share the notation ${\alpha }^{\beta }$ and disagree already at $2^{\omega }$, which is $\omega$ here. Ordinal ${\alpha }^{\beta }$ and cardinal ${\kappa }^{\lambda }$ are different operations that share one notation sets out the difference; FALSE: the ordinal $2^{\omega }$ is uncountable computes the value.