Cardinality of an arbitrary cartesian product Is it always possible to calculate the cardinality of the set of choices functions of a collection?
In others words, assuming AC (axiom of choice), we have that an arbitrary cartesian product is non-void, given that each of its sets is non-void. Is it possible to determinate its cardinality?
For example, given a collection of cardinals $\{\alpha_l\}_{l\in\lambda}$, how can we compute the cardinality of
$$\prod_{l\in\lambda}\alpha_l$$
knowing the cardinality of each $\alpha_l$ and $\lambda$?
An especific case of interest would be: how many choices functions are there in $P^2(\omega)\setminus\emptyset$? Does it depends of CH (continuum hypothesis) or GCH (generalized continuum hypotesis)?
 A: We know that $P(\omega)$ has $2^{\aleph_0}=\mathfrak{c}$ elements; and that $P^2(\omega)$ has $2^{\mathfrak{c}}$ elements. 
For simplicity, say $\mathfrak{d}=2^{\mathfrak{c}}$, and let $X=P^2(\omega)\setminus\varnothing$.
Then
$$\left|\prod_{A\in X}A\right| = \prod_{A\in X}|A| \leq \prod_{A\in X}\mathfrak{c} = \mathfrak{c}^{\mathfrak{d}}.$$
But we also have
$$\left|\prod_{A\in X}A\right| \geq \left|\prod_{B\in X,|B|=\mathfrak{c}}B\right| = \prod_{B\in X,|B|=\mathfrak{c}}|B| = \mathfrak{c}^{\mathfrak{d}}.$$
Also,
$$2^{\mathfrak{d}}\leq \mathfrak{c}^{\mathfrak{d}}\leq (\mathfrak{d}^+)^{\mathfrak{d}}\leq (2^{\mathfrak{d}})^{\mathfrak{d}} = 2^{\mathfrak{dd}} = 2^{\mathfrak{d}},$$
so we get equality throughout.
So we can certainly say that the number of choice functions on $X$ is $2^{\mathfrak{d}} = 2^{2^{\mathfrak{c}}}$.
In the absence of CH we don’t even know what $\mathfrak{c}$. Even if we do, we don’t know what $2^{\mathfrak{c}}$ is, and we don’t know what $2^{2^{\mathfrak{c}}}$ is.
But if we have GCH, then $\mathfrak{c}=\aleph_1$, $2^{\mathfrak{c}}=\aleph_2$, and $2^{2^{\mathfrak{c}}} = \aleph_3$. So if you allow GCH, the set you are asking about has cardinality $\aleph_3$.
