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

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

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.