 Diagonal argument for ultrapower cardinality

If $U$ is a kappa-complete nonprincipal ultrafilter on $\kappa$ and $|M|\ge \kappa$, then $|M^{\kappa }/U|>\kappa$. Otherwise enumerate representatives $f_{\alpha }$ for its classes and choose $g(i)\notin \{f_{\alpha }(i):\alpha <i\}$. Every tail of $\kappa$ belongs to $U$, so $g$ differs modulo $U$ from every representative. The small set is absent from a complete nonprincipal ultrafilter lemma supplies the tail property.