We denote by $|S|$ the number of elemets of a finite set $S$.
Lemma 1
Let $A$ be a local doman which is not a field.
Let $\mathfrak{m}$ be its maximal ideal.
Suppose $A$ satisfies the following conditions.
(1) $\mathfrak{m}$ is principal, i.e. there exists $t \in A$ such that $\mathfrak{m} = tA$.
(2) $0 = \bigcap_n \mathfrak{m}^n$
Then $A$ is a discrete valuation ring.
Proof:
Since $A$ is not a field, $t \neq 0$.
Let $x$ be a non-zero element of $\mathfrak{m}$.
By (2), there exists largest integer $n \geq 1$ such that $x \in \mathfrak{m}^n$.
We denote $n$ = ord($x$).
Since $x \in t^nA$, there exists $u \in A$ such that $x = t^n u$.
Clearly $u$ is invertible in $A$.
Hence $xA = t^nA$.
Let $I$ be a non-zero ideal of $A$ such that $I \subset \mathfrak{m}$.
Let $n$ = min{ord($x$);$x \in I - {0}$}.
There exists $y$ such that $n$ = ord($y$).
If $x \in I$ - {$0$}, ord($x$) $\geq n$.
Hence $x \in t^nA$.
Since $t^nA = yA$, $x \in yA$.
Hence $I = yA$.
Therefore $A$ is a principal ideal domain.
Hence $A$ is a discrete valuation ring.
QED
Lemma 2
Let $A$ be a Noetherian local doman which is not a field.
Let $\mathfrak{m}$ be its maximal ideal.
Suppose $\mathfrak{m}$ is principal.
Then $A$ is a discrete valuation ring.
Proof:
There exists $t \in A$ such that $\mathfrak{m} = tA$.
Since $A$ is not a field, $t \neq 0$.
Let $I = \bigcap_n \mathfrak{m}^n$.
By Lemma 1, it suffices to prove that $I = 0$.
Let $x \in I$.
There exists $y_n \in A$ such that $x = t^n y_n$ for each integer $n \geq 1$.
Since $x = t^n y_n = t^{n+1} y_{n+1}$, $y_n = ty_{n+1}$.
Hence $y_nA \subset y_{n+1}A$.
Since $A$ is Noetherian, $y_kA = y_{k+1}A$ for some integer $k$.
Hence $y_{k+1} = y_k u$.
Since $y_k = ty_{k+1} = tu y_k$.
Hence $(1 - tu)y_k = 0$.
Since $1 - tu$ is invertible, $y_k = 0$.
Hence $x = 0$.
Therefore $I = 0$.
QED
Lemma 3
Let $A$ be an integral domain.
Let $K$ be the field of fractions of A.
Let $\Omega$ be the set of maximal ideals of $A$.
Then $A = \bigcap_{P\in\Omega} A_P$, where we regard each $A_P$ as a subring of $K$.
Proof:
Let $x \in \bigcap_{P\in\Omega} A_P$.
Let $I$ = {$a \in A$; $ax \in A$}.
Clearly $I$ is an ideal.
Suppose $I \neq A$.
There exists a maximal ideal $P$ such that $I \subset P$.
Since $x \in A_P$, there exists $s \in A - P$ such that $sx \in A$.
Since $s \in I$, $s \in P$.
This is a contradiction.
Hence $I = A$.
Therefore $1 \in I$.
Hence $x \in A$.
Hence $\bigcap_{P\in\Omega} A_P \subset A$.
The other inclusion is clear.
QED
Lemma 4
Let $K$ be an algebraic number field.
Let $n = [K : \mathbb{Q}]$.
Let $B$ be the ring of algebraic integers in $K$.
Let $p$ be a prime number.
Suppose $pB = \alpha^nB$.
Then $\alpha B$ is a prime ideal of $B$ and $|B/\alpha B| = p$.
Proof:
Let $P$ be a prime ideal of $B$ lying over $p\mathbb{Z}$.
Since $\alpha^n \in P$, $\alpha \in P$.
Hence $pB \subset P^n$.
Hence $pB = P^n$.
Hence $\alpha^nB = P^n$.
Hence $\alpha B = P$.
Cearly $|B/\alpha B| = p$.
QED
Definition
Let $K$ be an algebraic number field.
Let $n = [K : \mathbb{Q}]$.
Let $\sigma_1,\dots,\sigma_n$ be distinct homomorphisms $K \rightarrow \mathbb{C}$.
Let $\omega_1,\dots,\omega_n$ be elements of $K$.
We denote det($\sigma_i(\omega_j)$) by $\Delta(\omega_1,\dots,\omega_n)$.
Lemma 5
Let $f(X) \in \mathbb{Z}[X]$ be a monic irreducible polynomial of degree $n$.
Let $\theta$ be a root of $f(X)$.
Let $A = \mathbb{Z}[\theta]$.
Let $p$ be a prime number.
Suppose $pA = \alpha^n A$, where $\alpha \in A$.
Then $\alpha A$ is a prime ideal of $A$ and $|A/\alpha A| = p$.
Proof:
Let $K$ be the field of fractions of $A$.
$1, \theta,\dots,\theta^{n-1}$ is a basis of $A$ as a free $\mathbb{Z}$-module.
Hence $\alpha,\alpha\theta,\dots,\alpha\theta^{n-1}$ is a basis of $\alpha A$ as a free $\mathbb{Z}$-module.
Hence $|A/\alpha A| = |\Delta(\alpha,\alpha\theta,\dots,\alpha\theta^{n-1})/\Delta(1, \theta,\dots,\theta^{n-1})| = |N_{K/\mathbb{Q}}(\alpha)|$.
Since $|N_{K/\mathbb{Q}}(\alpha)|$ is $p$ by Lemma 4, $A/\alpha A$ is a field.
Hence $\alpha A$ is a prime ideal of $A$.
QED
Proposition
Let $f(X) \in \mathbb{Z}[X]$ be a monic irreducible polynomial of degree $n$.
Let $\theta$ be a root of $f(X)$.
Let $A = \mathbb{Z}[\theta]$.
Let $p$ be a prime number.
Suppose the discriminant of $f(X)$ is a power of $p$.
Suppose $pA = \alpha^n A$, where $\alpha \in A$.
Then $A$ is integrally closed.
Proof:
Let $\Omega$ be the set of non-zero prime ideals of $A$.
By this, $\Omega$ is the set of maximal ideals of $A$.
By Lemma 3, $A = \bigcap_{P\in\Omega} A_P$.
Hence it suffices to prove that $A_P$ is integrally closed for each $P \in \Omega$.
Let $P \in \Omega$.
Suppose $P$ does not divide $p$.
Since the discriminant $d$ of $f(X)$ is a power of $p$, $P$ does not divide $d$.
Hence $A_P$ is integrally closed by my answers to this question.
Suppose $P$ divides $p$.
Since $P$ divides $\alpha$, By Lemma 5, $P = \alpha A$.
Hence $A_P$ is a discrete valuation ring by Lemma 2.
Hence $A_P$ is integrally closed.
QED
