 Solution

Because ${\kappa }_0<{\kappa }_1=\mathrm{crit}(j_1)$, the definition of the critical point of an elementary embedding immediately gives

To evaluate $j_0({\kappa }_1)$, first use the regularity of the measurable cardinal ${\kappa }_1$. Every function $f:{\kappa }_0\to {\kappa }_1$ has bounded range, so every ordinal below $j_0({\kappa }_1)$ lies below $j_0(\beta )$ for some $\beta <{\kappa }_1$. HenceThe strong-limit property of ${\kappa }_1$ gives ${\beta }^{{\kappa }_0}<{\kappa }_1$ for every $\beta <{\kappa }_1$. There are therefore fewer than ${\kappa }_1$ functions ${\kappa }_0\to \beta$, which implies $j_0(\beta )<{\kappa }_1$. On the other hand $j_0(\beta )\ge \beta$. Taking suprema yieldsthe two measurable cardinals under an ultrapower embedding formula.