Let $I \lhd R$ be a nonzero, proper ideal of a commutative Noetherian domain $R$. Why is it the case that we have $I^n \ne I^{n+1}$ for any $n \in \Bbb{N}$?
My thoughts: First note that if we have $I^n = I^{n + 1}$ for some $n \in \Bbb{N}$ then $$I^{n + 2} = I^{n + 1} I = I^n I = I^{n + 1} = I^{n}$$ so $I^{k} = I^n \quad \forall k \ge n$.
Also, as $R$ is Noetherian, $I$ is finitely generated, say by $\{a_1, \ldots, a_k\}$. Then $$I^n = \sum_{n_1 + \ldots + n_k = n} R {a_1}^{n_1} {a_2}^{n_2} \ldots {a_k}^{n_k}$$ and similarly we can find $I^{n + 1}$.
So $I^n = I^{n + 1} \;$ if and only if ${a_1}^{n_1} {a_2}^{n_2} \ldots {a_k}^{n_k} = \sum r {a_1}^{m_1} {a_2}^{m_2} \ldots {a_k}^{m_k} \;$ where $n_1 + \ldots + n_k = n$, and $m_1+ \ldots + m_k = n + 1$.
I'm not really sure how to proceed, thanks in advance for any help!
Just apply Nakayama's Lemma.
You've got that $I^nI=I^n$, so there exists an element $r\in R$ such that $1\equiv r\pmod I$ and $rI=0$. But since you're in a domain, $rI\neq \{0\}$.
You used the Noetherian condition to get that $I^n$ is finitely generated, and you'll find that the conditions $1\equiv r\pmod I$ and $rI=\{0\}$ in a domain are inconsistent with $I$ being a nonzero proper ideal.