Let $A=k[[X^2,X^3]]$. Statement: There exists no $\mathfrak p\in \mathrm{Spec}(A)$ s.t. $\mathfrak p^n=(X^3)$.
Is this correct proof?
Assume $\mathfrak p$ is such a prime. Then $A/\mathfrak p\cong k$ as both the generators must then be mapped to $0$ as if $x^3=0$ then $(x^2)^3=0$ so $x^2=0$ as $A/\mathfrak p$ is an integral domain (where $x^3$ and $x^2$ are the images of $X^3$ and $X^2$). So $\mathfrak p=(X^2,X^3)$ (the only max. ideal of $A$) which is a contradiction, as then $n=1$.