 Solution

For every $i\ge 0$, the inclusion $P^{i+1}\subseteq P^i$ is strict: equality, followed by multiplication by the inverse of $P^i$, would give $P=R$. Hence one can choosePut $J=(\pi )+P^{i+1}$. ThenThe classification just proved, applied modulo $P^{i+1}$, leaves no ideal strictly between consecutive powers. It follows thatFor $i=0$, the same argument says $(\pi )+P=R$, using $\pi \notin P$ and maximality.

The extension to every $r\ge 1$ follows by mathematical induction. Suppose $P^i=(\pi )+P^{i+r}$. Multiplication of the initial equality by $P^r$ givesSubstitution yieldsThusTaking $r=n-i$ for $0\le i<n$ givesThis is a principal ideal of the quotient ring $R/P^n$. Its zero ideal is generated by zero, so every ideal of this quotient ring is a principal ideal.