[proofplan]
We apply [Nakayama's lemma](/theorems/2168) to the finitely generated $\mathcal{O}_{C,p}$-module $M = \mathfrak{m}_p / (t)$. Since the residue $\bar t$ spans the one-dimensional $k$-vector space $\mathfrak{m}_p / \mathfrak{m}_p^2$, every element of $\mathfrak{m}_p$ equals a multiple of $t$ modulo $\mathfrak{m}_p^2$, which translates to $\mathfrak{m}_p \cdot M = M$. Nakayama then forces $M = 0$, i.e. $\mathfrak{m}_p = (t)$. The bulk of the proof is verifying the two hypotheses of Nakayama: that $\mathcal{O}_{C,p}$ is local (it is, by definition of the local ring at a point) and that $M$ is finitely generated (this uses that $\mathfrak{m}_p$ is finitely generated, which holds because $\mathcal{O}_{C,p}$ is a localisation of a Noetherian ring).
[/proofplan]
[step:Set up the module $M = \mathfrak{m}_p / (t)$ and verify it is finitely generated]
The local ring $\mathcal{O}_{C,p}$ at the smooth point $p$ of the curve $C$ is the localisation of the affine coordinate ring $A(C)$ at the maximal ideal corresponding to $p$. By [Hilbert's basis theorem](/theorems/???), $A(C)$ is Noetherian (being a quotient of the polynomial ring $k[X_1, \ldots, X_n]$), and localisation preserves the Noetherian property, so $\mathcal{O}_{C,p}$ is Noetherian. In particular, every ideal of $\mathcal{O}_{C,p}$ is finitely generated as an $\mathcal{O}_{C,p}$-module — including $\mathfrak{m}_p$.
Define
\begin{align*}
M := \mathfrak{m}_p / (t),
\end{align*}
the quotient $\mathcal{O}_{C,p}$-module. Since $\mathfrak{m}_p$ is finitely generated, so is its quotient $M$ (the images of generators of $\mathfrak{m}_p$ generate $M$).
[/step]
[step:Use the one-dimensionality of $\mathfrak{m}_p / \mathfrak{m}_p^2$ to write $\mathfrak{m}_p = (t) + \mathfrak{m}_p^2$]
The quotient $\mathfrak{m}_p / \mathfrak{m}_p^2$ is naturally a vector space over the residue field $k = \mathcal{O}_{C,p} / \mathfrak{m}_p$ (the multiplication of $\mathcal{O}_{C,p}$ on $\mathfrak{m}_p / \mathfrak{m}_p^2$ factors through the quotient by $\mathfrak{m}_p$ because $\mathfrak{m}_p \cdot \mathfrak{m}_p \subseteq \mathfrak{m}_p^2$). By hypothesis,
\begin{align*}
\dim_k \mathfrak{m}_p / \mathfrak{m}_p^2 = 1, \qquad \bar t = t + \mathfrak{m}_p^2 \neq 0.
\end{align*}
A nonzero element of a one-dimensional $k$-vector space is a basis. Hence every element $\bar x \in \mathfrak{m}_p / \mathfrak{m}_p^2$ can be written uniquely as $\bar x = \bar c\, \bar t$ for some $\bar c \in k$.
Lift this to $\mathcal{O}_{C,p}$. For every $x \in \mathfrak{m}_p$, choose any $c \in \mathcal{O}_{C,p}$ representing $\bar c \in k = \mathcal{O}_{C,p} / \mathfrak{m}_p$. Then $x - c\, t \in \mathfrak{m}_p^2$, so
\begin{align*}
x \in c\, t + \mathfrak{m}_p^2 \subseteq (t) + \mathfrak{m}_p^2.
\end{align*}
Since this holds for every $x \in \mathfrak{m}_p$, we conclude
\begin{align*}
\mathfrak{m}_p = (t) + \mathfrak{m}_p^2.
\end{align*}
[/step]
[step:Translate $\mathfrak{m}_p = (t) + \mathfrak{m}_p^2$ into $\mathfrak{m}_p \cdot M = M$]
Reduce the equality $\mathfrak{m}_p = (t) + \mathfrak{m}_p^2$ modulo $(t)$: as submodules of $\mathfrak{m}_p$, we have
\begin{align*}
\mathfrak{m}_p / (t) = \big((t) + \mathfrak{m}_p^2\big) / (t).
\end{align*}
The right side is the image of $\mathfrak{m}_p^2$ in the quotient $\mathfrak{m}_p / (t) = M$. Now $\mathfrak{m}_p^2$ is the $\mathcal{O}_{C,p}$-submodule of $\mathfrak{m}_p$ generated by products $ab$ with $a, b \in \mathfrak{m}_p$; equivalently, $\mathfrak{m}_p^2 = \mathfrak{m}_p \cdot \mathfrak{m}_p$. Its image in $M$ is therefore
\begin{align*}
\mathfrak{m}_p \cdot \big(\mathfrak{m}_p / (t)\big) = \mathfrak{m}_p \cdot M.
\end{align*}
Combining,
\begin{align*}
M = \mathfrak{m}_p \cdot M.
\end{align*}
[guided]
The translation from "$\mathfrak{m}_p = (t) + \mathfrak{m}_p^2$" to "$\mathfrak{m}_p \cdot M = M$" is the key bookkeeping move that lets us apply Nakayama. Why is this the right reduction? Nakayama operates on a module $M$ via the relation $\mathfrak{m} M = M$, so we need to engineer a module where this relation holds.
The right module is $M = \mathfrak{m}_p / (t)$. The submodule $(t)$ is killed in $M$ by construction — that is the entire point of dividing by $(t)$. So any decomposition $\mathfrak{m}_p = (t) + (\text{something})$ becomes $M = 0 + (\text{image of something in } M) = (\text{image of something in } M)$ inside $M$.
Plugging in: $\mathfrak{m}_p = (t) + \mathfrak{m}_p^2$ becomes $M = \overline{\mathfrak{m}_p^2}$ in $M$, where the bar denotes image in $M$. The image of $\mathfrak{m}_p^2 = \mathfrak{m}_p \cdot \mathfrak{m}_p$ in $M = \mathfrak{m}_p / (t)$ is $\mathfrak{m}_p \cdot M$ — the $\mathfrak{m}_p$-submodule of $M$ generated by the images of $\mathfrak{m}_p$. Hence $M = \mathfrak{m}_p M$, the Nakayama hypothesis.
This is a standard technique for invoking Nakayama: when you want to show two submodules $N \subseteq N'$ of a module are equal, where the "extra" generators of $N'$ lie in $\mathfrak{m} N'$, set $M = N' / N$ and show $\mathfrak{m} M = M$. Nakayama then gives $M = 0$, i.e. $N = N'$.
[/guided]
[/step]
[step:Apply Nakayama's lemma to $M$ to conclude $\mathfrak{m}_p = (t)$]
We apply [Nakayama's lemma](/theorems/2168) to the $\mathcal{O}_{C,p}$-module $M$. Verify the hypotheses:
\begin{itemize}
\item $\mathcal{O}_{C,p}$ is a local ring with maximal ideal $\mathfrak{m}_p$. This is the [characterisation of $\mathcal{O}_{C,p}$ as a local ring](/theorems/2128), which holds because $\mathcal{O}_{C,p}$ is the localisation of $A(C)$ at a maximal ideal.
\item $M = \mathfrak{m}_p / (t)$ is finitely generated as an $\mathcal{O}_{C,p}$-module, by Step 1.
\item $\mathfrak{m}_p \cdot M = M$, by Step 3.
\end{itemize}
Nakayama's lemma yields $M = 0$, i.e.
\begin{align*}
\mathfrak{m}_p / (t) = 0 \quad \Longleftrightarrow \quad \mathfrak{m}_p = (t).
\end{align*}
This completes the proof.
[/step]
