= Solution
Taking norms in
$$
p\mathcal O_K=Q_1\cdots Q_s
$$
gives
$$
p^n=N(p\mathcal O_K)=\prod_{i=1}^sN(Q_i)=p^s,
$$
so $s=n$.
Suppose for contradiction that $\mathcal O_K=\mathbb Z[\alpha]$, and let $f$ be the degree-$n$ minimal polynomial of $\alpha$. Part (a) identifies the prime ideals above $p$ with the distinct irreducible factors of $\overline f$. Since there are $n$ distinct prime ideals, all with norm $p$, the norm formula forces $\overline f$ to have $n$ distinct irreducible factors of degree one. But $\mathbb F_p[X]$ has only the $p$ distinct monic linear polynomials
$$
X-a\qquad(a\in\mathbb F_p).
$$
This is impossible when $n>p$. The <completely split small-prime obstruction to monogenicity> therefore gives
$$
\boxed{n>p\quad\Longrightarrow\quad
\mathcal O_K\ne\mathbb Z[\alpha]\text{ for every }\alpha\in\mathcal O_K.}
$$
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