 Solution

An uncountable cardinal $\kappa$ is measurable when it carries a nonprincipal ultrafilter $U$ that is $\kappa$-complete. For an inaccessible $\lambda >\kappa$, the cardinal $\kappa$ is 1-strong when there is an elementary embeddinginto a transitive model, with critical point $\kappa$ and $V_{\kappa +1}\subseteq M$.

The fundamental theorem on measurable cardinals constructs from $U$ the well-founded ultrapower and its ultrapower embedding $j:V_{\lambda }\to M$, whose critical point is $\kappa$. The embedding fixes $V_{\kappa }$. If $A\in V_{\kappa +1}$, then $A\subseteq V_{\kappa }$, and elementarity givesBoth $j(A)$ and $V_{\kappa }$ belong to the transitive target, so $A\in M$. Thus $V_{\kappa +1}\subseteq M$, proving that every measurable cardinal is 1-strong.