[proofplan] We use that, for the full modular group, every cusp is equivalent to the cusp represented by $\infty$. Thus the cusp-form condition is the vanishing of the value at $\infty$, and we translate that value into the local coordinate $q=e^{2\pi i z}$. A modular form $f \in M_k$ is holomorphic at $\infty$ exactly when it is represented near $q=0$ by a [holomorphic function](/page/Holomorphic%20Function) whose Taylor coefficients are the $q$-expansion coefficients of $f$. The value of $f$ at the cusp $\infty$ is therefore the value of this holomorphic extension at $q=0$, which is its constant Taylor coefficient. [/proofplan] [step:Represent a modular form near $\infty$ in the $q$-coordinate] Fix $f \in M_k$. Let $\mathbb{D} := \{q \in \mathbb{C}: |q|<1\}$ denote the open unit disk. Since $f$ is holomorphic at the cusp $\infty$, there is a holomorphic map \begin{align*} F_f:\mathbb{D} &\to \mathbb{C} \end{align*} such that, for every $z$ in the upper half-plane with $q=e^{2\pi i z}$ sufficiently close to $0$, \begin{align*} F_f(q)=f(z). \end{align*} By definition of the $q$-expansion, the Taylor expansion of $F_f$ at $0$ is \begin{align*} F_f(q)=\sum_{n=0}^{\infty} a_n(f)q^n. \end{align*} [/step] [step:Identify the value at the cusp with the constant coefficient] The value of $f$ at the cusp $\infty$ is defined by \begin{align*} f(\infty):=F_f(0). \end{align*} Evaluating the Taylor expansion of $F_f$ at $q=0$ gives \begin{align*} F_f(0)=\sum_{n=0}^{\infty} a_n(f)0^n=a_0(f), \end{align*} because every term with $n \geq 1$ vanishes at $q=0$. Hence \begin{align*} f(\infty)=0 \iff a_0(f)=0. \end{align*} [/step] [step:Translate the cusp condition into the stated set equality] For the full modular group, the set of cusps has a single equivalence class, represented by $\infty$. Hence the definition of cusp form gives: $f \in S_k$ if and only if $f \in M_k$ and $f$ vanishes at that representative cusp, namely $f(\infty)=0$. From the previous step, this is equivalent to $f \in M_k$ and $a_0(f)=0$. Therefore \begin{align*} S_k=\{f \in M_k : a_0(f)=0\}. \end{align*} This proves the theorem. [/step]