$Ass(M)$ is finite

Let $ M$ be a f.g. module over a Noetherian ring. I’m trying to fill in the details in the proof of the fact that $ Ass(M)$ is finite. It should follow from Proposition 3.7 and Lemma 3.6 in Eisenbud.

First, consider a filtration $ 0=M_0\subset\dots\subset M_n=M$ where $ M_{i+1}/M_{i}\simeq R/P_i$ for some prime ideals $ P_i$ (Proposition 3.7 in Eisenbud). Let us prove by induction on $ n$ that $ Ass(M_n)\subset \{P_0,\dots, P_{n-1}\}$ .

First of all, I have problems with the base case: if $ n=1$ , the chain looks like $ 0=M_0\subset M_1=M$ , and $ M_1\simeq M_1/M_0\simeq R/P_0$ for some prime ideal $ P_0$ . So $ P_0$ is an associated prime of $ M$ , so $ \{P_0\}\subset Ass(M)$ . Why does the reverse inclusion hold?

Assume the result holds for all $ k<n$ . Consider the exact sequence $ 0\to M_{n-1}\to M_n\to M_n/M_{n-1}\to 0$ . By lemma 3.6, $ $ Ass(M_n)\subset Ass(M_{n-1})\cup Ass(M_n/M_{n-1})$ $

By hypothesis $ Ass(M_{n-1})\subset \{P_0,\dots,P_{n-2}\}$ . But what is $ Ass$ of the quotient, and why?