\proclaim Lemma.
If $f\in M_k(N)$ and
  $a_n(f)=0$ for $n\leq {kN\over 12}\cdot \prod_{p|N}(1+{1 \over p}) $, then
$f=0$.

\smallskip\noindent{\bf Proof.}\  
We must show that the composite map 
$$M_k(N)\hookrightarrow{\bf C}[[q]]\rightarrow{\bf C}[[q]]/(q^{r+1})$$
is injective.  Because ${\bf C}$ is a flat ${\bf Z}$-module, it suffices 
to show that the map $\Phi:M_k(N;{\bf Z})\rightarrow{\bf Z}[[q]]/(q^{r+1})$ 
is injective.  Suppose $\Phi(f)=0$, and let~$p$ be a prime number.
Then $a_n(f)=0$ for $n\leq r$, hence plainly $a_n(f)\equiv 0\pmod{p}$
for any such~$n$.  By Theorem~1 of STURM, it follows that $f\equiv 0\pmod{p}$. 
Repeating this argument shows that the coefficients of~$f$ 
are divisible by all primes~$p$, i.e., they are~$0$.

Let $\mu(N)=N\cdot \prod_{p|N}(1+{1 \over p})$ denote the index of
$\Gamma_0(N)$ in ${\rm SL}_2({\bf Z})$.