Integrally Closed domain and Principal Ideal 
Let $R$ be an integrally closed local domain. Suppose there is a $y\in I^n$ such that $yI^n=I^{2n}$ for some $n$. I would like to prove that $I^n=(y)$.

Source: The above question comes from the proof of Lemma F on page 449 of this paper.
A proof is given, but i cannot understand it, I seem to be missing something very simple it seems.
I asked this question earlier and somebody downvoted, so i thought about it a little more, but still i feel i am missing something. 

In such questions the Proposition 2.4 from Atiyah and Macdonald, Introduction to Commutative Algebra, is a kind of universal tool. (See also here.)
In the notation of the book, set $M=I^n/y$, $\mathfrak a= R$. Note that $M^2=M$ within the field of fractions of $R$. Now let $x\in M$ and define $\phi:M\to M$ by $\phi(a)=ax$. Then $\phi(M)\subseteq M$, and therefore there are $a_i\in R$ such that $$\phi^n+a_1\phi^{n-1}+\cdots+a_n=0.$$ Thus $a(x^n+a_1x^{n-1}+\cdots+a_n)=0$ for all $a\in M$. In particular, $x^n+a_1x^{n-1}+\cdots+a_n=0$, so $x$ is integral over $R$ hence $x\in R$, and we are done.
