= Solution
For a weight-$k$ <cusp form>, the <invariant norm of a modular form> $y^{k/2}|f(z)|$, with $y=\operatorname{Im}z$, is unchanged by the <modular group>: $\operatorname{Im}(\gamma z)=y/|cz+d|^2$ and $|f(\gamma z)|=|cz+d|^k|f(z)|$ cancel exactly.
Every orbit meets the <standard fundamental domain of the modular group> $\mathcal F=\{z:|\operatorname{Re}z|\leq1/2,\ |z|\geq1\}$, where $y\geq\sqrt3/2$. The <Fourier expansion of a modular form> at the cusp has $f(q)=q h(q)$, 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
$$
\boxed{B:=\sup_{z\in\mathfrak H}y^{k/2}|f(z)|<\infty.}
$$
For every $y>0$, <Fourier coefficient> extraction gives
$$
a_n(f)=e^{2\pi ny}\int_0^1f(x+iy)e^{-2\pi inx}\,dx,
\qquad |a_n(f)|\leq B e^{2\pi ny}y^{-k/2}.
$$
For $n\geq1$, minimize the last factor by taking $y=k/(4\pi n)$. This proves the <Fourier coefficient bound for a cusp form>
$$
\boxed{|a_n(f)|\leq B\left(\frac{4\pi e}{k}\right)^{k/2}n^{k/2}<Cn^{k/2},
\quad C=B\left(\frac{4\pi e}{k}\right)^{k/2}+1.}
$$
The assertion concerns positive indices; the constant coefficient vanishes because $f$ is a <cusp form>.
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