Find the associated primes of $(x_0) \cdot (x_0, x_1) \cdot \dots \cdot (x_0, \dots, x_r)$ in $k[x_0, \dots x_r]$ I'm studying Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry, and I'm having some trouble with the following problem (Exercise 3.9 in the book):

Let $k$ be a field, and let $I$ be the ideal $(x_0) \cdot (x_0, x_1) \cdot \dots \cdot (x_0, \dots, x_r)$ of $k[x_0, \dots x_r]$. Show that the associated primes of $I$ are $(x_0), (x_0, x_1), \dots, (x_0, \dots, x_r)$.

The back of the book contains the hint

Do induction on $r$, inverting $x_r$, and using Theorem 3.1.

Theorem 3.1, among other things, states that the associated primes of $R[U^{-1}]$ correspond to the associated primes of $R$ that are disjoint from $U$. Based on this, it seems like the inductive step ought to involve proving something like this:

Let $R_i$ be the ring
  $$k[x_0, x_1, \dots, x_r, x_r^{-1}, x_{r-1}^{-1} \dots, x_{i+1}^{-1}],$$
  and let $I_i$ be the ideal
  $$(x_0) \cdot (x_0, x_1) \cdot \dots \cdot (x_0, \dots, x_i)$$
  of $R_i$. Assume that the associated primes of $I_i$ are $(x_0), (x_0, x_1), \dots, (x_0, \dots, x_i)$. Then, the associated primes of $I_{i+1}$ are $(x_0), (x_0, x_1), \dots, (x_0, \dots, x_{i+1})$.

By Theorem 3.1, it suffices to show that the only associated prime of $I_{i+1}$ that contains $x_{i+1}$ is $(x_0, \dots, x_{i+1})$. However, I'm not sure how to show this.
Is this what Eisenbud meant with his hint? If so, how do I finish the proof?
 A: Let $R = k[x_0,\dots,x_r]$, $P_n = (x_0,x_1,\dots,x_n)$ and define $I_n = P_0 P_1 \cdots P_n$. Clearly $I_0 = P_0$ is a prime ideal and so the set of associated primes of $R/I_0$ is just $\left\{P_0\right\}$. Now, suppose by induction that the associated primes of $R/I_n$ are precisely $\left\{P_0,P_1\dots,P_n\right\}$. Consider $I_{n+1}$. Since $I_{n+1}R_{x_{n+1}} = I_n R_{x_{n+1}}$, it follows that all $P_0,\dots,P_n$ are associated primes of $R/I_{n+1}$. Moreover, $P_{n+1}$ is precisely $I_{n+1} : x_0 x_1 \dots x_n$, and so $P_{n+1}$ is an associated prime of $R / I_{n+1}$ as well. Now let us show that there are no other associated primes. Towards that end, notice that the associated primes of $I_{n+1}$ are ideals of the form $(x_{j_1},\dots,x_{j_\ell})$, with $j_i \le n+1$, since the smallest ring that contains $I_{n+1}$ is $k[x_0,\dots,x_{n+1}]$. If $Q$ is an associated prime different than $P_0,\dots,P_{n+1}$, then one of its generators has to be $x_{n+1}$ (otherwise localizing at $x_{n+1}$, $Q$ must be one of the $P_0,\dots,P_n$). Since $Q \supset I_{n+1}$, $Q$ must contain one of the $P_0, \dots\, P_{n+1}$. Since $x_{n+1} \in Q$ the only possibility is $Q \supset P_{n+1}$, from which it follows that $Q = P_{n+1}$, contradiction and proof completed.
In fact, it can be shown that
\begin{align}
(x_0)(x_0,x_1)\cdots(x_0,\dots,x_r) = (x_0) \cap (x_0,x_1)^2 \cap (x_0,x_1,x_2)^3 \cap \cdots \cap (x_0,\dots,x_r)^{r+1},\end{align} where the right-hand-side is a primary decomposition of the the left-hand-side.
