 Solution

For a weight-$k$ cusp form, the invariant norm of a modular form $y^{k/2}|f(z)|$, with $y=\mathrm{Im}z$, is unchanged by the modular group: $\mathrm{Im}(\gamma z)=y/|cz+d{|}^2$ and $|f(\gamma z)|=|cz+d{|}^k|f(z)|$ cancel exactly.

Every orbit meets the standard fundamental domain of the modular group $\mathcal{F}=\{z:|\mathrm{Re}z|\le 1/2,|z|\ge 1\}$, where $y\ge \sqrt{3}/2$. The Fourier expansion of a modular form at the cusp has $f(q)=qh(q)$, with $h$ holomorphic near zero, since $f$ is a cusp form. Thus $|f(x+iy)|=O(e^{-2\pi y})$ uniformly in $x$ as $y\to \infty$, and $y^{k/2}|f|$ tends to zero there. On the remaining compact part of $\mathcal{F}$ it is bounded by continuity. Invariance therefore gives

For every $y>0$, Fourier coefficient extraction givesFor $n\ge 1$, minimize the last factor by taking $y=k/(4\pi n)$. This proves the Fourier coefficient bound for a cusp formThe assertion concerns positive indices; the constant coefficient vanishes because $f$ is a cusp form.