= Solution
First use the fact that a <small set is absent from a complete nonprincipal ultrafilter>. Indeed, if $A\subseteq\kappa$ and $|A|<\kappa$, then every $\kappa\setminus\{\alpha\}$, $\alpha\in A$, belongs to $U$ by nonprincipality. <Kappa-completeness> gives $\kappa\setminus A\in U$, so $A\notin U$. In particular every final segment $\{i:i>\alpha\}$ belongs to $U$.
Suppose for contradiction that the <ultrapower> has at most $\kappa$ elements. List representatives $f_\alpha:\kappa\to M$ for all its classes, indexed by $\alpha<\kappa$; repetitions are allowed. At coordinate $i<\kappa$, fewer than $\kappa$ values occur among $f_\alpha(i)$ with $\alpha<i$. Since $|M|\geq\kappa$, choose
$$
g(i)\in M\setminus\{f_\alpha(i):\alpha<i\}.
$$
This <diagonal argument for ultrapower cardinality> uses the <axiom of choice>. For each fixed $\alpha$, the functions $g$ and $f_\alpha$ disagree throughout the final segment $i>\alpha$, which belongs to $U$. Their equality set therefore cannot belong to $U$, and $[g]\ne[f_\alpha]$.
This contradicts the assumed enumeration of the <ultrapower>. Hence
$$
\boxed{|M^\kappa/U|>\kappa.}
$$
The proof uses only that each ordinal $i<\kappa$ has <cardinality> below $\kappa$; it does not require a separate assumption of regularity.
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