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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.

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 $|M|\ge \kappa$, chooseThis 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. HenceThe proof uses only that each ordinal $i<\kappa$ has cardinality below $\kappa$; it does not require a separate assumption of regularity.

Let $U$ be a nonprincipal $\kappa$-complete ultrafilter on the measurable cardinal $\kappa$, and fix $\lambda <\kappa$. Suppose for a contradiction that $2^{\lambda }\ge \kappa$. Choose distinct subsets $A_{\alpha }\subseteq \lambda$ for $\alpha <\kappa$. For each $\xi <\lambda$, the ultrafilter property chooses exactly one ofas a member $H_{\xi }$ of $U$. Since $\lambda <\kappa$, $\kappa$-completeness gives $H={\bigcap }_{\xi <\lambda }{H}_{\xi }\in U$.

Any two indices in $H$ give subsets having the same membership decision at every $\xi <\lambda$, so they give the same $A_{\alpha }$. The chosen subsets were distinct, hence $|H|\le 1$. This contradicts the fact that a small set is absent from a complete nonprincipal ultrafilter. Therefore $2^{\lambda }<\kappa$ for every $\lambda <\kappa$, which is precisely the strong limit cardinal condition. This is the measurable cardinal is a strong limit cardinal argument.