Lemma 1 Let $A$ be a Noetherian local domain which is not a field. Suppose its maximal ideal $\mathfrak{m}$ is principal. Then $\bigcap_n \mathfrak{m}^n = 0$.
Proof: Let $\mathfrak{m} = tA$. Let $x \in \bigcap_n \mathfrak{m}^n$. Suppose $x \neq 0$. There exists $y_n \in A$ for every $n$ such that $x = t^ny_n$. Then $t^ny_n = t^{n+1}y_{n+1}$. Hence $y_n = ty_{n+1}$. Hence $y_nA \subset y_{n+1}A$. Since $A$ is Noetherian, there exists $k$ such that $y_kA = y_{k+1}A$. Hence there exists $u \in A$ such that $y_{k+1} = uy_k$. Since $y_k = ty_{k+1}$, $y_k = uty_k$. Hence $(1 - ut)y_k = 0$. Since $t \in \mathfrak{m}$, $1 - ut$ is invertible. Hence $y_k = 0$. Hence $x = t^ky_k = 0$. This is a contradiction. QED
Lemma 2 Let $A$ be a Noetherian local domain which is not a field. Suppose its maximal ideal $\mathfrak{m}$ is principal. Then $A$ is a discrete valuation ring.
Proof: Suppose $\mathfrak{m} = tA$. By Lemma 1, $\bigcap_n \mathfrak{m}^n = 0$. Let $I$ be a non-zero ideal of $A$. There exists $n$ such that $I \subset \mathfrak{m}^n$ but not $I \subset \mathfrak{m}^{n+1}$. Since $\mathfrak{m}^n = t^nA$, $It^{-n} \subset A$. Suppose $It^{-n} \neq A$. Then $It^{-n} \subset \mathfrak{m}$. Hence $I \subset \mathfrak{m}^{n+1}$. This is a contradictin. Hence $I = t^nA$. QED
Lemma 3 Let $A$ be an integral domain. Let $I$ be an ideal of $A$. Suppose $I$ is invertble. Then $I$ is a finitely generated projective $A$-module.
Proof: Since $II^{-1} = A$, there exist $a_1,\dots,a_n \in I$ and $b_1,\dots,b_n \in I^{-1}$ such that $\sum_i a_ib_i = 1$. Let $f_i:I\rightarrow A$ be the $A$-linear map defined by $f_i(x) = b_ix$. Let $L$ be a free $A$-module with a basis $e_1,\dots,e_n$. Let $g:L \rightarrow I$ be the $A$-linear map defined by $g(e_i) = a_i$. Let $f:I \rightarrow L$ be the $A$-linear map defined by $f(x) = \sum_i f_i(x)e_i = \sum_i b_ixe_i$. Since $gf(x) = \sum_i g(b_ixe_i) = \sum_i b_ia_ix = x$ for every $x \in I$, $gf = 1$. Hence $I$ is isomorphic to a direct summand of $L$. Hence $I$ is a finitely generated projective $A$-module. QED
Lemma 4 Let $A$ be a local ring. Let $M$ be a finitely generated projective $A$-module. Then $M$ is a finitely generated free $A$-module.
Proof: Let $\mathfrak{m}$ be the maximal ideal of $A$. Let $k = A/\mathfrak{m}$. Since $M$ is finitely generated, dim$_k M\otimes_A k$ is finite. Let $a_1,\dots,a_n$ be elements of $M$ such that $\{a_1\otimes 1,\dots,a_n\otimes 1\}$ is a basis of $M\otimes_A k$ over $k$. By Nakayama's lemma, $a_1,\dots,a_n$ generates $M$ over $A$. Let $L$ be a free $A$-module with a basis $\{e_1,\dots,e_n\}$. Let $f:L\rightarrow M$ be the $A$-linear map such that $f(e_i) = a_i (i = 1,\dots,n)$. Let $K$ be the kernel of $f$. Then we get the following exact sequence.
$0 \rightarrow K \rightarrow L \rightarrow M \rightarrow 0$
Then the following sequence is exact by the well known theorem of homological algebra.
Tor$_1(M, k) \rightarrow K\otimes_A k \rightarrow L\otimes_A k \rightarrow M\otimes_A k \rightarrow 0$
Since $M$ is projective, Tor$_1(M, k) = 0$. Since $L\otimes_A k \rightarrow M\otimes_A k$ is an isomorphism, $K\otimes_A k = 0$. Since $M$ is projective, $K$ is a direct summand of $L$. Hence $K$ is finitely generated. Hence $K = 0$ by Nakayama's lemma. QED
Lemma 5 Let $A$ be a Noetherian local domain. Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal. Let $K$ be the field of fractions of $A$. Let $\mathfrak{m}^{-1} = \{x \in K; x\mathfrak{m} ⊂ A\}$. Then $\mathfrak{m}^{-1} \neq A$.
Proof: Let $a \neq 0$ be an element of $\mathfrak{m}$. By the assumption, Supp$(A/aA) = \{\mathfrak{m}\}$. Since Ass$(A/aA) \subset$ Supp($A/aA)$, Ass$(A/aA) = \{\mathfrak{m}\}$. Hence there exists $b \in A$ such that $b \in A - aA$ and $\mathfrak{m}b \subset aA$. Since $\mathfrak{m}(b/a) \subset A$, $b/a \in \mathfrak{m}^{-1}$. Since $b \in A - aA$, $b/a \in K - A$. QED
Lemma 6 Let $A$ be an integral domain. Let $K$ be the field of fractions of $A$. Let $M \neq 0$ be a finitely generated $A$-submodule of $K$. Let $x \in K$ be such that $xM \subset M$. Then $x$ is integral over $A$.
Proof: Let $\omega_1,\dots,\omega_n$ be generators of $M$ over $A$. Let $x\omega_i = \sum_j a_{i,j} \omega_j$. Then $x$ is a root of the characteristic polynomial of the matrix $(a_{ij})$. QED
Lemma 7 Let $A$ be an integrally closed Noetherian local domain. Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal. Then $\mathfrak{m}$ is invertible.
Proof: Let $K$ be the field of fractions of $A$. Let $a \neq 0$ be an element of $\mathfrak{m}$. Let $\mathfrak{m}^{-1} = \{x \in K; x\mathfrak{m} \subset A\}$. Since $\mathfrak{m} \subset \mathfrak{m}\mathfrak{m}^{-1} \subset A$, $\mathfrak{m}\mathfrak{m}^{-1} = \mathfrak{m}$ or $\mathfrak{m}\mathfrak{m}^{-1} = A$. Suppose $\mathfrak{m}\mathfrak{m}^{-1} = \mathfrak{m}$. Since $\mathfrak{m}$ is finitely generated, every element of $\mathfrak{m}^{-1}$ is integral over $A$ by Lemma 6. Since $A$ is integrally closed, $\mathfrak{m}^{-1} \subset A$. This is a contradiction by Lemma 5. Hence $\mathfrak{m}\mathfrak{m}^{-1} = A$ and therefore $\mathfrak{m}$ is invertible. QED
Lemma 8 Let $A$ be an integral domain. Let $L$ be a finitely generated free A-module. Let $M$ be a finitely generated free A-submodule of L. Then rank$_A M \le$ rank$_A L$.
Proof: Let $K$ be the field of fractions of $A$. Since $K$ is a flat $A$-module, the canonical homomorphism $M\otimes_A K \rightarrow L\otimes_A K$ is injective. Hence we are done. QED
Theorem Let $A$ be an integrally closed Noetherian local. Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal. Then $A$ is a discrete valuation ring.
Proof: By Lemma 7, $\mathfrak{m}$ is invertible. By Lemma 3, $\mathfrak{m}$ is projective over A. By Lemma 4, $\mathfrak{m}$ is a finitely generated free module $A$. By Lemma 8, $\mathfrak{m}$ is principal. Hence $A$ is a discrete valuation ring by Lemma 2. QED