I would like to prove that, for a modular form $f$ that is cuspidal and of weight $k$, we have $a_f(n) \ll n^{k/2}.$
However all the proofs I find (e.g. in Gelbart) are a bit intricate. It does come directly from the definition and easy consideration but there is something less "trivial" than I expected about the proof. Are there short or enlightening proof of it?
This is known as the "Hecke bound".
Fix a weight $k$ holomorphic cuspidal modular form $f = \sum_{n \geq 1} a(n) q^n$ on a congruence subgroup $\Gamma \subseteq \mathrm{SL}(2, \mathbb{Z})$.
First, one can compute that $\mathrm{Im}(z)^{k/2} \lvert f(z) \rvert$ is invariant under the action of $\Gamma$. As $f$ is cuspidal, $\mathrm{Im}(z)^{k/2} \lvert f(z) \rvert$ is bounded at the various cusps, and thus bounded on $\mathbb{H}$ as it's invariant under the action of $\Gamma$. Denote the bound by $$ \mathrm{Im}(z)^{k/2} \lvert f(z) \rvert \leq C. $$
We can pick out the $n$th Fourier coefficient via $$ \lvert a(n) \rvert = \left \lvert \int_0^1 f(x + iy) e^{- 2 \pi i n (x + iy)} dx \right \rvert \leq C y^{-k/2} \int_0^1 e^{2 \pi n y} dx \leq C y^{-k/2} e^{2 \pi n y}. $$ Each choice of $y$ gives a bound for $\lvert a(n) \rvert$. This is optimized when $y = 1/n$, giving the bound $$ \lvert a(n) \rvert \leq C e^{2 \pi} n^{k/2}.$$