= Solution
Write $D=(2\pi i)^{-1}d/dz=q\,d/dq$ and use the <Serre derivative> $D_kf=Df-(k/12)E_2f$. Translation invariance of $f$ and the <Eisenstein series of weight two> gives $D_kf(z+1)=D_kf(z)$.
Differentiate the weight-$k$ transformation $f(-1/z)=z^kf(z)$. The <chain rule> yields
$$
Df(-1/z)=z^{k+2}Df(z)+\frac{k}{2\pi i}z^{k+1}f(z).
$$
The assumed transformation of the <Eisenstein series of weight two> gives
$$
\frac{k}{12}E_2(-1/z)f(-1/z)
=\frac{k}{12}z^{k+2}E_2(z)f(z)+\frac{k}{2\pi i}z^{k+1}f(z).
$$
Subtracting cancels the extra term. Hence $D_kf(-1/z)=z^{k+2}D_kf(z)$. Since $S$ and $T=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ generate the <modular group>, these two transformations prove the weight-$k+2$ law for all its elements. Holomorphy on the <complex upper half-plane> follows from the formula.
For the <Fourier expansion of a modular form> $f=\sum_{n\geq0}a_nq^n$, differentiation gives $Df=\sum_{n\geq1}na_nq^n$. The product of the convergent series for $E_2$ and $f$ likewise has no negative powers. Thus $D_kf$ is <holomorphic at a cusp> at infinity, and therefore at every cusp of the <modular group>. Its constant coefficient is $-ka_0/12$. Since $k>0$, it vanishes exactly when $a_0=0$. Consequently
$$
\boxed{D_kf\in M_{k+2}(\Gamma(1)),\qquad
D_kf\in S_{k+2}(\Gamma(1))\iff f\in S_k(\Gamma(1)).}
$$
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