Showing that the hyperintegers are uncountable In class, we constructed the hyperintegers as follows:
Let $N$ be a normal model of the natural number with domain $\mathbb{N}$ in the language $\{0, 1, +, \cdot, <, =\} $. Also let $F$ be a fixed nonprincipal ultrafilter on $\omega$. Then we have $N^*$ as the ultrapower $N^\omega /F$ with domain $\mathbb{N}^{\omega} / F$ . 
Now I need to show that $N^*$ is uncountable (this means that it's cardinality is $2^{\aleph_0}$ I guess). 
The domain of $N^*$ are the equivalence classes of functions $\{f_F : f \mbox{ a function from }\mathbb{N} \to \mathbb{N}\}$
Further on, we have $N^{\sharp}$ which is the equivalence classes of all the constant functions. 
This is my intuition: I know that $N$ and $N^{\sharp}$ are isomporphic (hence have the same cardinality, which is $\aleph_0$ since $N$ is a model for the natural numbers). Furthermore I know that $N^{\sharp}$ is elementary equivalent to $N^*$. But $N^*$ has an extra 'copy' of $N^\sharp$ on top of it, the hyperintegers. So then the cardinalty of $N^*$ should be $2^{\aleph_0}$ (does that even exist? And is that uncountable?). I hope someone can help me with a formal proof.

A further variation on this theme is to choose a fixed infinite hyperinteger $H$ and note that the partial map $f\colon{}^\ast\mathbb{N}\to\mathbb{R}$ given by $f(n)=\text{st}(\frac{n}{H})$ whenever this is defined, is surjective.
