= Solution
The <height of a prime ideal> is the supremum of the lengths of strict chains of <prime ideals> ending at $P$:
$$
\operatorname{ht}P=\sup\{d:P_0\subsetneq P_1\subsetneq\cdots\subsetneq P_d=P\}
=\dim R_P.
$$
The last equality follows from the <prime ideal correspondence for localization>. Here is a length-growth proof of the <Krull height theorem> that makes the bound explicit.
We first establish a <prime-chain lower bound for local length>. For a <Noetherian local ring> $(B,\mathfrak m)$ admitting a <prime chain> of length $d$, put $H_B(t)=\operatorname{length}(B/\mathfrak m^t)$. Then
$$
H_B(t)\geq c\,t^d\qquad(t\gg0)
$$
for some $c>0$. For $d=0$, use $H_B(t)\geq1$. For $d>0$, quotient by the first <prime ideal> of the chain; this only decreases $H_B$ and reduces us to a <local integral domain> $B$ with
$$
0\subsetneq P_1\subsetneq\cdots\subsetneq P_d=\mathfrak m.
$$
Take $0\ne a\in P_1$ and put $C=B/(a)$. It has a <prime chain> of length $d-1$, so induction bounds $H_C(t)$ below by a positive multiple of $t^{d-1}$.
The <Artin-Rees lemma> applied to $(a)\subseteq B$ gives, for a fixed integer $c_0\geq1$,
$$
(a)\cap\mathfrak m^t\subseteq a\mathfrak m^{t-c_0}\qquad(t\geq c_0).
$$
Because $a$ is a <non-zero-divisor>, the first term in the <exact sequence>
$$
0\longrightarrow (a)/((a)\cap\mathfrak m^t)
\longrightarrow B/\mathfrak m^t
\longrightarrow B/((a)+\mathfrak m^t)\longrightarrow0
$$
has length at least $H_B(t-c_0)$. Hence
$$
H_B(t)\geq H_B(t-c_0)+H_C(t).
$$
Iterate this inequality about $t/(2c_0)$ times, while its arguments remain at least $t/2$. This gives the desired positive multiple of $t^d$. The particular <Artin-Rees lemma> inclusion used here follows directly from finite generation: the <Rees algebra> $\bigoplus_{j\geq0}\mathfrak m^jt^j$ is <Noetherian>, by the <Hilbert basis theorem>, and its graded submodule $\bigoplus_j((a)\cap\mathfrak m^j)t^j$ has finitely many homogeneous generators. A bound on their degrees gives $c_0$.
Now localize at the given minimal <prime ideal> $P$. Write $B=R_P$, $\mathfrak m=PR_P$, and $J=IR_P$. The only <prime ideal> of $B$ containing $J$ is $\mathfrak m$, so $\sqrt J=\mathfrak m$. Since $\mathfrak m$ is finitely generated, some $\mathfrak m^q\subseteq J$. Thus $B/J$ has finite <module length>, say $L$. The <ideal> $J$ is generated by at most $n$ elements.
For $n>0$, products of these generators give a surjection from $\binom{j+n-1}{n-1}$ copies of $B/J$ onto $J^j/J^{j+1}$. Summing the lengths gives the <generator bound for primary-ideal length>
$$
\operatorname{length}(B/J^t)
\leq L\sum_{j=0}^{t-1}\binom{j+n-1}{n-1}
=L\binom{t+n-1}{n}=O(t^n).
$$
Since $J\subseteq\mathfrak m$, we have $J^t\subseteq\mathfrak m^t$, and therefore
$$
H_B(t)\leq\operatorname{length}(B/J^t)=O(t^n).
$$
The <prime-chain lower bound for local length> now excludes any chain of length greater than $n$. \b[Consequently]
$$
\boxed{\operatorname{ht}P=\dim R_P\leq n.}
$$
For $n=0$, minimality over the zero <ideal> says that $P$ is a minimal <prime ideal>, so its height is zero. This also handles that boundary case.
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