Every prime power ideal in a Noetherian Ring of dimension one can be written uniquely as a power of a prime.

It suffices to show that the chain $$mathfrak psupseteqmathfrak p^2supseteqmathfrak p^3 supseteq cdots$$ is strictly descending.

Suppose instead that $mathfrak p^{n+1}=mathfrak p^n$ for some positive integer $n$.

Let $M=mathfrak p^n$, regarded as an $A$-module.

Since $A$ is a domain and $mathfrak pne 0$, we get $Mne 0$.

Since $A$ is Noetherian, $M$ is finitely generated.

From $mathfrak p^{n+1}=mathfrak p^n$ we get $mathfrak pM=M$, hence by Nakayama's lemma

$;;;;$ https://en.wikipedia.org/wiki/Nakayama%27s_lemma#Statement

we get $aM=0$ for some $ain A$ with $aequiv 1;(text{mod};mathfrak p)$.

But then $amathfrak p^n=0$, contradiction, since $ane 0$,$;mathfrak p^nne 0$, and $A$ is a domain.