[proofplan] First we use the previously established modularity theorem for Hecke operators to obtain that $T_n f$ is again a modular form whenever $f$ is a modular form. The only additional condition needed for cusp forms is vanishing of the constant Fourier coefficient at the cusp $\infty$. We compute the constant coefficient of $T_n f$ from the Hecke coefficient formula and observe that it is a scalar multiple of the constant coefficient of $f$. [/proofplan] [step:Use the modularity theorem for Hecke operators] Let $\mathbb{H} := \{z \in \mathbb{C} : \operatorname{Im}(z) > 0\}$ denote the complex upper half-plane. Let $f \in M_k(\mathrm{SL}_2(\mathbb{Z}))$ be a modular form of weight $k$ for the full modular group. By the standard modularity theorem for Hecke operators, already established before this result, the function \begin{align*} T_n f: \mathbb{H} &\to \mathbb{C} \end{align*} is again a modular form of weight $k$ for $\mathrm{SL}_2(\mathbb{Z})$; that is, \begin{align*} T_n f \in M_k(\mathrm{SL}_2(\mathbb{Z})). \end{align*} Thus \begin{align*} T_n\bigl(M_k(\mathrm{SL}_2(\mathbb{Z}))\bigr) \subseteq M_k(\mathrm{SL}_2(\mathbb{Z})). \end{align*} Here the cited result is the preceding theorem in the development: [Hecke operators preserve modular forms](/theorems/4244). [/step] [step:Compute the constant Fourier coefficient of $T_n f$] Now assume $f \in S_k(\mathrm{SL}_2(\mathbb{Z}))$. Let $z \in \mathbb{H}$ be the upper half-plane variable, define $q := e^{2\pi i z}$, and let $a_m(f) \in \mathbb{C}$ denote the $m$th Fourier coefficient of $f$ at the cusp $\infty$ for each integer $m \geq 0$. Since $f$ is a cusp form, it has Fourier expansion at the cusp $\infty$ \begin{align*} f(z)=\sum_{m=1}^{\infty} a_m(f) q^m, \end{align*} so its constant Fourier coefficient satisfies \begin{align*} a_0(f)=0. \end{align*} The Fourier coefficient formula for the weight-$k$ Hecke operator says that, for every integer $m \geq 0$, \begin{align*} a_m(T_n f)=\sum_{d \mid \gcd(m,n)} d^{k-1} a_{mn/d^2}(f), \end{align*} with the convention that $\gcd(0,n)=n$. Setting $m=0$, the divisors $d$ are exactly the positive divisors of $n$, and $mn/d^2=0$. Therefore \begin{align*} a_0(T_n f) &=\sum_{d \mid n} d^{k-1} a_0(f) \\ &=\left(\sum_{d \mid n} d^{k-1}\right)a_0(f) \\ &=0. \end{align*} [/step] [step:Conclude that $T_n f$ is a cusp form] From the first step, $T_n f \in M_k(\mathrm{SL}_2(\mathbb{Z}))$. From the constant coefficient computation, \begin{align*} a_0(T_n f)=0. \end{align*} For modular forms on $\mathrm{SL}_2(\mathbb{Z})$, vanishing of the constant Fourier coefficient at the cusp $\infty$ is exactly the cusp condition. Hence \begin{align*} T_n f \in S_k(\mathrm{SL}_2(\mathbb{Z})). \end{align*} Since $f \in S_k(\mathrm{SL}_2(\mathbb{Z}))$ was arbitrary, we have \begin{align*} T_n\bigl(S_k(\mathrm{SL}_2(\mathbb{Z}))\bigr) \subseteq S_k(\mathrm{SL}_2(\mathbb{Z})). \end{align*} Combining this with the preservation of modular forms proves the theorem. [/step]