 Solution

The height $\mathrm{ht}\mathfrak{p}$ is the supremum of lengths of strict chains of primes ending at $\mathfrak{p}$. The Krull principal ideal theorem says that in a Noetherian ring every prime minimal over a principal proper ideal has height at most one.

Induct on $n$. If $\mathfrak{p}$ is minimal over $I=(x_1,\dots ,x_n)$, choose a prime $\mathfrak{q}\subseteq \mathfrak{p}$ minimal over $(x_1,\dots ,x_{n-1})$ and localize appropriately. In $R/\mathfrak{q}$, the prime $\mathfrak{p}/\mathfrak{q}$ is minimal over the principal ideal generated by $x_n$, so its relative height is at most one. Induction gives $\mathrm{ht}\mathfrak{q}\le n-1$, hence