Here $\alpha,\beta$ are Von Neumman ordinals, and $\alpha^{\beta}$ is the ordinal exponentiation. It seems to me that the cardinality $\#(\alpha^{\beta})=\#\alpha^{\#\beta}$. But my textbook does not mention it. So is that true? May I please ask how can I see the fact?
EDIT: Knowing it is not true. May I please ask if it works for finite case? If it does not work, may I please ask for a way to determine the cardinality of ordinal exponentiation? Thanks!
The basic rule is:
Theorem. Given $\alpha > 1, \beta > 0,$ and at least one of $\alpha, \beta$ is infinite. $\text{card}(\alpha^\beta) = \max \lbrace \text{card}(\alpha), \text{card}(\beta) \rbrace$.
For the proof, we will use
Lemma 1. If $\alpha > 1$, $\alpha^\beta$ is a strictly increasing function of $\beta$.
Proof of Lemma 1: This follows from
$\alpha^{\beta+1} = \alpha^\beta \cdot \alpha > \alpha^\beta$
If $\beta$ is a limit ordinal, for any $\gamma < \beta$, $\alpha^\beta \ge \alpha^{\gamma+1} > \alpha^\gamma$. (End proof of lemma 1)
So for $\alpha > 1, \beta > 0$, we have $\alpha^\beta > \alpha$. Further, we have $\alpha^\beta \ge \beta$, since $\alpha^\beta$ is the $\beta$th ordinal in the range of $f(\gamma) = \alpha^\gamma$. So $\alpha^\beta \ge \max \{\alpha,\beta\}$, and therefore $\text{card}(\alpha^\beta) \ge \text{card}(\max \{\alpha,\beta\}) = \max\{\text{card}(\alpha),\text{card}(\beta)\}$.
For the other direction, we will use the fact that $\alpha^\beta$ can be viewed as the set of functions from $\beta$ to $\alpha$ with finite support. We can partition this set by the number of elements of $\beta$ that map to something other than $0$. The size of the partition with $n$ such elements is the number of ways of choosing $n$ distinct elements of $\beta$, multiplied with the number of ways of choosing an $n$-tuple of $\alpha \setminus \{0\}$. The former is clearly no more than $\text{card}(\beta^n)$, and the latter no more than $\text{card}(\alpha^n)$. So
$$\text{card}(\alpha^\beta) \le \bigcup_n \text{card}(\alpha^n) \text{card}(\beta^n) = \bigcup_n \max \{\text{card}(\alpha),\text{card}(\beta)\} = \max \{\text{card}(\alpha),\text{card}(\beta)\} \aleph_0$$ $$ = \max \{\text{card}(\alpha),\text{card}(\beta)\}$$
with the last equality due to the assumption that at least one of $\alpha,\beta$ is infinite. (End of proof of theorem)
So a full list of rules for cardinality of ordinal exponentiation:
$\alpha^0 = 1$.
For $\beta > 0$, $0^\beta = 0$.
$1^\beta = 1$.
If $\alpha,\beta$ are finite, $\text{card}(\alpha^\beta) = \alpha^\beta$.
Otherwise, $\text{card}(\alpha^\beta) = \max \lbrace \text{card}(\alpha), \text{card}(\beta) \rbrace$.
In particular, you asked whether $\text{card}(\alpha^\beta) = \text{card}(\alpha)^{\text{card}(\beta)}$ in the finite case. This follows immediately from $\text{card}(n) = n$ for finite $n$.