[proofplan] Choose a holomorphic coordinate chart centred at $p$ and identify germs of holomorphic functions near $p$ with germs of holomorphic functions near $0 \in \mathbb{C}^n$ by composition with the chart inverse. Taylor expansion identifies the latter germ ring with the ring of convergent power series $\mathbb{C}\{z_1,\ldots,z_n\}$. The local-ring assertion then follows from the elementary fact that a holomorphic germ is invertible exactly when its value at the base point is nonzero; hence the non-units are precisely the germs vanishing at $p$, and this set is the unique maximal ideal. [/proofplan]

[step:Identify the stalk with holomorphic germs at the origin in coordinates] Let $n$ denote the complex dimension of $X$ at $p$. Choose a holomorphic coordinate chart \begin{align*} \varphi: U &\to V \subseteq \mathbb{C}^n \end{align*} where $U \subseteq X$ is an open neighbourhood of $p$, $V \subseteq \mathbb{C}^n$ is open, and $\varphi(p) = 0$. Write \begin{align*} z_i: U &\to \mathbb{C} \\ q &\mapsto \pi_i(\varphi(q)) \end{align*} for $1 \leq i \leq n$, where $\pi_i: \mathbb{C}^n \to \mathbb{C}$ is the $i$-th coordinate projection. Define \begin{align*} \Phi: \mathcal{O}_{X,p} &\to \mathcal{O}_{\mathbb{C}^n,0} \\ f_p &\mapsto (f \circ \varphi^{-1})_0, \end{align*} where $f: W \to \mathbb{C}$ is any holomorphic representative of the germ $f_p$ on an open neighbourhood $W \subseteq U$ of $p$, and $(f \circ \varphi^{-1})_0$ denotes the germ at $0$ of the holomorphic map \begin{align*} f \circ \varphi^{-1}: \varphi(W) &\to \mathbb{C}. \end{align*} This is well-defined: if $f$ and $g$ represent the same germ at $p$, then there is an open neighbourhood $W_0 \subseteq W$ of $p$ on which $f=g$, so $f \circ \varphi^{-1}=g \circ \varphi^{-1}$ on the open neighbourhood $\varphi(W_0)$ of $0$. The map $\Phi$ is a ring homomorphism because composition with $\varphi^{-1}$ preserves pointwise addition and multiplication. Its inverse is \begin{align*} \Psi: \mathcal{O}_{\mathbb{C}^n,0} &\to \mathcal{O}_{X,p} \\ h_0 &\mapsto (h \circ \varphi)_p, \end{align*} where $h: A \to \mathbb{C}$ is a holomorphic representative on an open neighbourhood $A \subseteq V$ of $0$. The same germ argument proves that $\Psi$ is well-defined, and the identities $\Psi \circ \Phi = \operatorname{id}_{\mathcal{O}_{X,p}}$ and $\Phi \circ \Psi = \operatorname{id}_{\mathcal{O}_{\mathbb{C}^n,0}}$ follow after restricting representatives to smaller neighbourhoods. Thus $\Phi$ is a ring isomorphism. [/step]

[step:Identify holomorphic germs at the origin with convergent power series]Let $\mathbb{C}\{z_1,\ldots,z_n\}$ denote the ring of power series \begin{align*} \sum_{\alpha \in \mathbb{N}^n} a_\alpha z^\alpha, \end{align*} where $a_\alpha \in \mathbb{C}$, $z^\alpha = z_1^{\alpha_1}\cdots z_n^{\alpha_n}$, and the series converges absolutely on some polydisc \begin{align*} P_r = \{w=(w_1,\ldots,w_n) \in \mathbb{C}^n : |w_i| < r \text{ for every } 1 \leq i \leq n\} \end{align*} for some real number $r>0$. Define \begin{align*} T: \mathcal{O}_{\mathbb{C}^n,0} &\to \mathbb{C}\{z_1,\ldots,z_n\} \\ h_0 &\mapsto \sum_{\alpha \in \mathbb{N}^n} \frac{\partial^\alpha h(0)}{\alpha!} z^\alpha, \end{align*} where $h: A \to \mathbb{C}$ is a holomorphic representative on an open neighbourhood $A \subseteq \mathbb{C}^n$ of $0$, $\alpha! = \alpha_1!\cdots \alpha_n!$, and $\partial^\alpha h(0)$ denotes the mixed complex partial derivative of multi-index $\alpha$ at $0$. By the several-variable holomorphic Taylor expansion theorem, this Taylor series converges on some polydisc contained in $A$ and represents $h$ there. Therefore $T$ is well-defined on germs. Conversely, every element of $\mathbb{C}\{z_1,\ldots,z_n\}$ converges on some polydisc $P_r$ and defines a holomorphic function $P_r \to \mathbb{C}$, hence a germ in $\mathcal{O}_{\mathbb{C}^n,0}$. This construction is inverse to $T$, again by uniqueness of Taylor coefficients for holomorphic functions. Addition and multiplication of convergent power series agree with addition and multiplication of the represented holomorphic germs, so $T$ is a ring isomorphism. Composing $T$ with $\Phi$ gives the coordinate-dependent ring isomorphism \begin{align*} \mathcal{O}_{X,p} \cong \mathbb{C}\{z_1,\ldots,z_n\}. \end{align*}[/step]

[guided]The coordinate chart reduces the statement from the manifold $X$ to ordinary holomorphic functions near $0 \in \mathbb{C}^n$. Once we are in $\mathbb{C}^n$, the correct algebraic object is the ring of convergent power series, not the larger formal power series ring. Let $\mathbb{C}\{z_1,\ldots,z_n\}$ be the ring of all series \begin{align*} \sum_{\alpha \in \mathbb{N}^n} a_\alpha z^\alpha \end{align*} with complex coefficients $a_\alpha \in \mathbb{C}$ that converge absolutely on at least one polydisc around $0$. Here the multi-index notation means \begin{align*} z^\alpha = z_1^{\alpha_1}\cdots z_n^{\alpha_n}. \end{align*} For a holomorphic germ $h_0 \in \mathcal{O}_{\mathbb{C}^n,0}$, choose a representative \begin{align*} h: A &\to \mathbb{C}, \end{align*} where $A \subseteq \mathbb{C}^n$ is an open neighbourhood of $0$. The several-variable Taylor theorem gives a convergent expansion \begin{align*} h(w)=\sum_{\alpha \in \mathbb{N}^n} \frac{\partial^\alpha h(0)}{\alpha!} w^\alpha \end{align*} on some polydisc $P_r \subseteq A$. Thus the rule \begin{align*} T: \mathcal{O}_{\mathbb{C}^n,0} &\to \mathbb{C}\{z_1,\ldots,z_n\} \\ h_0 &\mapsto \sum_{\alpha \in \mathbb{N}^n} \frac{\partial^\alpha h(0)}{\alpha!} z^\alpha \end{align*} assigns a convergent power series to each germ. This assignment is independent of the representative. If two representatives agree as germs at $0$, then they agree on some neighbourhood of $0$, so all their derivatives at $0$ agree. Conversely, a convergent power series defines a holomorphic function on a sufficiently small polydisc, hence defines a germ at $0$. Uniqueness of Taylor coefficients shows that these two constructions are inverse to each other. Since Taylor expansion respects sums and products of holomorphic functions, $T$ is a ring isomorphism. Combining $T$ with the coordinate isomorphism from the previous step gives \begin{align*} \mathcal{O}_{X,p} \cong \mathbb{C}\{z_1,\ldots,z_n\}. \end{align*}[/guided]

[step:Characterize the units by nonvanishing at the base point]Define the evaluation map \begin{align*} \operatorname{ev}_p: \mathcal{O}_{X,p} &\to \mathbb{C} \\ f_p &\mapsto f(p), \end{align*} where $f: W \to \mathbb{C}$ is any holomorphic representative on an open neighbourhood $W \subseteq X$ of $p$. This is well-defined because representatives of the same germ agree on some neighbourhood of $p$. We prove that $f_p \in \mathcal{O}_{X,p}$ is a unit if and only if $\operatorname{ev}_p(f_p) \neq 0$. If $f_p$ is a unit, choose $g_p \in \mathcal{O}_{X,p}$ with $f_p g_p = 1$. Applying $\operatorname{ev}_p$ gives \begin{align*} f(p)g(p)=1, \end{align*} so $f(p) \neq 0$. Conversely, suppose $f(p) \neq 0$. Choose a holomorphic representative \begin{align*} f: W &\to \mathbb{C} \end{align*} on an open neighbourhood $W \subseteq X$ of $p$. Since $f$ is continuous and $f(p) \neq 0$, there is an open neighbourhood $W_1 \subseteq W$ of $p$ such that $f(q) \neq 0$ for every $q \in W_1$. Define \begin{align*} g: W_1 &\to \mathbb{C} \\ q &\mapsto \frac{1}{f(q)}. \end{align*} The reciprocal of a nonvanishing holomorphic function is holomorphic, so $g$ is holomorphic. Hence $g_p \in \mathcal{O}_{X,p}$ and $f_p g_p = 1$. Therefore $f_p$ is a unit exactly when $f(p) \neq 0$.[/step]

[guided]We now determine which germs have multiplicative inverses in the stalk. First define the value of a germ at $p$ by \begin{align*} \operatorname{ev}_p: \mathcal{O}_{X,p} &\to \mathbb{C} \\ f_p &\mapsto f(p), \end{align*} where $f: W \to \mathbb{C}$ is a representative on an open neighbourhood $W$ of $p$. This is independent of the representative: two representatives of the same germ agree on some neighbourhood of $p$, and therefore have the same value at $p$. Suppose first that $f_p$ is a unit. Then there is a germ $g_p \in \mathcal{O}_{X,p}$ such that \begin{align*} f_p g_p = 1. \end{align*} Evaluating this equality at $p$ gives \begin{align*} f(p)g(p)=1. \end{align*} A complex number that has a multiplicative inverse is nonzero, so $f(p)\neq 0$. Now suppose $f(p)\neq 0$. Choose a holomorphic representative \begin{align*} f: W &\to \mathbb{C} \end{align*} defined on an open neighbourhood $W \subseteq X$ of $p$. The point of using the value $f(p)$ is that nonvanishing is an open condition for a continuous function. Since $f$ is holomorphic, it is continuous, and since $f(p)\neq 0$, there exists an open neighbourhood $W_1 \subseteq W$ of $p$ such that $f(q)\neq 0$ for every $q \in W_1$. On this smaller neighbourhood define \begin{align*} g: W_1 &\to \mathbb{C} \\ q &\mapsto \frac{1}{f(q)}. \end{align*} The function $g$ is holomorphic because it is the reciprocal of a nonvanishing holomorphic function. Its germ $g_p$ satisfies \begin{align*} f_p g_p = 1. \end{align*} Therefore $f_p$ is a unit. We have proved the equivalence: a germ in $\mathcal{O}_{X,p}$ is invertible exactly when its value at $p$ is nonzero.[/guided]

[step:Show the vanishing germs form the unique maximal ideal] Let \begin{align*} \mathfrak{m}_p = \ker(\operatorname{ev}_p) = \{ f_p \in \mathcal{O}_{X,p} : f(p)=0 \}. \end{align*} Since $\operatorname{ev}_p$ is a ring homomorphism, $\mathfrak{m}_p$ is an ideal of $\mathcal{O}_{X,p}$. The map $\operatorname{ev}_p$ is surjective because every constant complex number $c \in \mathbb{C}$ is represented by the constant holomorphic map \begin{align*} c_X: X &\to \mathbb{C} \\ q &\mapsto c. \end{align*} Thus \begin{align*} \mathcal{O}_{X,p}/\mathfrak{m}_p \cong \mathbb{C}, \end{align*} so $\mathfrak{m}_p$ is maximal. It remains to prove uniqueness. Let $\mathfrak{a} \subseteq \mathcal{O}_{X,p}$ be any proper ideal. If $\mathfrak{a}$ contained a unit $u_p$, then $1=u_p^{-1}u_p$ would lie in $\mathfrak{a}$, contradicting propriety. Hence every element of $\mathfrak{a}$ is a non-unit. By the unit characterization above, every non-unit belongs to $\mathfrak{m}_p$, so $\mathfrak{a} \subseteq \mathfrak{m}_p$. In particular every maximal ideal is contained in $\mathfrak{m}_p$, and since $\mathfrak{m}_p$ itself is maximal, every maximal ideal equals $\mathfrak{m}_p$. Therefore $\mathcal{O}_{X,p}$ is a local ring with unique maximal ideal $\mathfrak{m}_p$. [/step]