Let $A$ be a commutative local noetherian ring and let $I$ be a proper ideal of $A$. Prove that $\bigcap_{n=1}^\infty I^n = 0$.
The first thing I tried was to see that $$\displaystyle\bigcap_{n = 1}^\infty I^n \subseteq \displaystyle\bigcap_{n = 2}^\infty I^n \subseteq \displaystyle\bigcap_{n = 3}^\infty I^n \subseteq \cdots$$ but since of these sets are equal, that was no avail. Then I realized that if I let $K := \bigcap_{n=1}^\infty I^n$, and if I show that $\mathfrak{m}K = K$, where $\mathfrak{m}$ is the unique maximal ideal of $A$, then by the fact that $\mathfrak{m}$ is the unique maixmal ideal we get that $K = 0$. However, I am having a little difficulty showing that $K \subseteq \mathfrak{m}K$. Let $k \in K$. Then for any $n \in \mathbb N$, there exists some $m \in \mathbb N$ and elements in $I$ such that $l = a_{11} \cdots a_{1n} + \cdots + a_{m1} \cdots a_{mn} \in I^n$. Since $I$ is a proper ideal, we have $I \subseteq \mathfrak{m}$. We thus have that $k \in \mathfrak{m}I^n$ for each $n \in \mathbb N$. It seems like I am missking something obvious, but how can I proceed to show that $k \in \mathfrak{m}K$? Where is Noetherian part used?
We can use that $$ \bigcap_{n=1}^\infty I^n \subseteq \bigcap_{n=1}^\infty \mathfrak{m}^n = \{ 0\},$$ where the later is the Krull intersection theorem and the former from the fact that $\mathfrak{m}$ is the unique maximal ideal and $I$ is a proper ideal.
The proof of the Krull intersection theorem is based on Nakayama's lemma as $$ \mathfrak{m}\cdot \bigcap_{n=1}^\infty \mathfrak{m}^n = \bigcap_{n=2}^\infty \mathfrak{m}^n = \bigcap_{n=1}^\infty \mathfrak{m}^n$$ and $\bigcap_{n=1}^\infty \mathfrak{m}^n$ is finitely generated (as $R$ is noetherian) we get that $\bigcap_{n=1}^\infty \mathfrak{m}^n = \{0\}$.