Is $\alpha^{\beta}$ a cardinal? Let $\alpha , \beta$ cardinals. Is $\alpha ^{\beta}$, defined as the set of all functions $f:\beta\to \alpha$, a cardinal?
I do this question because an autor of a text book says that the exponentiation of cardinals $\alpha$ raised to $\beta$ is defined as the cardinal of the set $\alpha ^{\beta}$. Then I ask my question, isn't it always $\alpha ^{\beta}$ a cardinal?
 A: Cardinal exponentiation is defined by saying that the cardinal number $\kappa^\lambda$ is the cardinality of the set of functions from $\lambda$ to $\kappa$. If you also denote that set of functions by $\kappa^\lambda$, then of course the symbol $\kappa^\lambda$ is ambiguous: it can be either the set of functions from $\lambda$ to $\kappa$ or the cardinality of that set. These are, however, two different things, even though many people denote them by the same symbol.
That is why some of us prefer to write ${}^XY$ for the set of functions from $X$ to $Y$: in that notation
$$\kappa^\lambda=\left|{}^\lambda\kappa\right|\;,$$
where $\kappa^\lambda$ is unambiguously the cardinal number that is the cardinal exponential $\kappa$ raised to the power $\lambda$, and ${}^\lambda\kappa$ is unambiguously the set of functions from $\lambda$ to $\kappa$.
A: If $\alpha$ and $\beta$ are cardinals, then $\alpha^\beta$ is the cardinality of the set of all functions from a set of cardinality $\beta$ to a set of cardinality $\alpha$.
If $A$ and $B$ are sets, then $A^B$ is the set of all functions from $B$ into $A$.
According to one way of making set theory precise, each cardinal is also a set, and so $\alpha^\beta$ could mean either of two different things.
However, that particular way of being precise is one option, and not necessarily sacred.
A: Consider for example $\alpha\cdot\beta$. That is the cardinal of the set $\alpha\times\beta$. But that set is not a cardinal itself.
In $\sf ZFC$ cardinals are ordinals, and $\alpha\times\beta$ is not an ordinal. You can observe that all its elements are ordered pairs, where as every ordinal includes $\varnothing$, which is not an ordered pair.
Even more so, $\alpha\cdot\beta=\max\{\alpha,\beta\}$. Certainly we have that neither is a set of ordered pairs. It's just the cardinal of one.
Generally we define cardinal arithmetic to be "the cardinal of some resulting set". In particular we take the cardinal $\alpha^\beta$ to be the cardinal of $\{f\subseteq\alpha\times\beta\mid f\text{ is a function}\}$, or generally $\{f\subseteq A\times B\mid f\text{ is a function}\}$, where $|A|=\alpha,|B|=\beta$ (this definition holds for a broader context, where cardinals are not necessarily initial ordinals).
One caveat about notation, $\alpha^\beta$ also denotes ordinal exponentiation sometimes. That's a whole other thing, and they give strangely different results. $2^\omega=\omega$ as ordinal exponentiation, but $2^\omega>\omega$ as cardinal exponentiation.
