The factor ring of the n-th power of a maximal ideal is local.

Question

Let $M$ be a maximal ideal in a commutative ring $R$ with identity and $n$ is a positive integer, then the ring $R/M^n$ has a unique prime ideal and therefore is local. It is easy to see that unique prime ideal implies local. But how to show there is a unique prime ideal?

Answer 1

Primes in $R/M^n$ correspond to primes, that contain $M^n$. If a prime $P$ contains $M^n$ - by the prime property - it contains $M$, hence $P=M$ by maximality of $M$.