 Solution

A cusp form is holomorphic in the upper half-plane by definition, so it is meromorphic there. In the reverse direction suppose $f$ is meromorphic andOn a small closed disk around an interior point, $y$ is bounded above and away from zero. Hence $|f|$ is bounded there by a fixed constant. A pole of positive order would make $|f|$ unbounded along points approaching its center, a contradiction. Every possible pole is therefore removable, and $f$ is holomorphic on $\mathbb{H}$. Meromorphicity alone would not establish this conclusion; the bounded invariant norm is being used here.