A few weeks ago, I had such homework exercise: Prove that if $U$ is any non-principal ultrafilter on $\Bbb{N}$, then the ultrapower $\Bbb{N^N}/U$ has cardinality continuum. ($\mathbb{N}$ stands for natural numbers.)
Solution shown by my professor was different than mine, so I want You to verify my solution: However, it works only assuming Continnum Hypothesis.
My solution:
It is obvious, that it has power less or equal continuum, because $\mathbb{N^N}$ has power continuum.
Assume it is countable. So, we can index its elements by natural numbers. Let $(a_1,a_2,a_3,\cdots)$ - sequence of all element of that ultrapower $\mathbb{N^N}/U$. They can be written as sequences of natural numbers. Pick sequence of elements of ultrafilter $(u_1,u_2,u_3,\cdots)$ such that every coordinate belongs only to finitely many of these elements (it is possible bcs $U$ is nonprincipal).
Now construct a Cantor diagonal element (lets call it $C$) such that $C$ differs from $a_i$ exactly on coordinates belonging to $U_i$. It is possible, because every coordinate belongs only to finitely many $U_j$'s and there are infinitely many naturals.
$C$ belongs to our ultrapower $\mathbb{N^N}/U$ and is different from every element $\{a1,a2,a3,\cdots\}$ because Ui belongs to ultrafilter.
Hence, we have contradiction. That can't have cardinality $\aleph_0$. More, assuming Continuum Hypothesis, it has power continuum.
Is it ok?