Assume $A$ is noetherian, $I\subseteq A$ is an ideal satisfying $I^s=0$, and $I^i/I^{i+1}$ is finite free $(A/I)$-module $\forall 0\le i\le s-1$. How can i prove $\mathrm{Ass}_A(A)=\mathrm{Ass}_A(A/I^i)$ for every $i$?
This question came out when I study the proof of Theorem 24.5 in Matsumura's Commutative Ring Theory, page 188. By taking appropriate principal open set we can reduce the situation over and over, and finally get the conditions in my question. Then the proof said "It's then easy to see that, If $x_1, \dots, x_n\in A$ is an $A/P$-sequence, it's also an $A$-sequence". At that time it's not easy to see for me, and what I've tried is that by induction I can reduce to the case $n=1$, and this comes to my question.
We will use the following result a few times.
(*): For a short exact sequence of finitely generated $A$-modules $0 \to N \to M \to L \to 0$, one has
(i) $\operatorname{ass} M \subset \operatorname{ass} N \cup \operatorname{ass} L$ and
(ii) $\operatorname{ass} N \subset \operatorname{ass} M$.
For each $i$, we have the following exact sequence $$ 0 \to I^i/I^{i+1} \to A/I^{i+1} \to A/I^i \to 0. $$
Using the assumption $I^{i}/I^{i+1}$ is a free $A/I$ module, by (*.ii), we have $\operatorname{ass} A/I \subset \operatorname{ass} A/I^{i+1}$ for any $0 \le i \le s-1$. In particular, when $i = s-1$, since $I^s = 0$, one has $\operatorname{ass} A/I \subset \operatorname{ass} A$.
By applying (*,i) to each exact sequence, one was $$ \operatorname{ass} A = \operatorname{ass} A/I^s \subset \operatorname{ass} A/I \cup \operatorname{ass} A/I^{s-1} \subset \operatorname{ass} A/I \cup \operatorname{ass} A/I^{s-2} \subset \cdots \subset \operatorname{ass} A/I \cup \operatorname{ass} A/I = \operatorname{ass} A/I.$$ Thus, one has $\operatorname{ass} A = \operatorname{ass} A/I$ and from this, for all $0 \le i \le s-1$, by the above inclusions, one has $\operatorname{ass} A/I^{i+1} \subset \operatorname{ass} A/I$. We already showed the other inclusion, so these two sets are equal.