Proof. In the proof of Lemma 15.12.1 we have seen that $A^ h$ is a filtered colimit of étale $A$-algebras $B$ such that $A/I \to B/IB$ is an isomorphism and we have seen that $I^ h = IA^ h$. As an étale ring map is flat (Algebra, Lemma 10.142.3) we conclude that $A \to A^ h$ is flat by Algebra, Lemma 10.38.3. Since each $A \to B$ is flat we find that the maps $A/I^ n \to B/I^ nB$ are isomorphisms as well (for example by Algebra, Lemma 10.100.3). Taking the colimit we find that $A/I^ n = A^ h/I^ nA^ h$ as desired. $\square$
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