 Solution

For a nonzero Hecke eigenform that is a cusp form, $c(1)\ne 0$: otherwise part (b) would make all its positive Fourier coefficients zero, and cuspidality already gives $c(0)=0$, forcing $f=0$. Since $T_w(1)=I$, $\lambda (1)=1$. Part (b) gives multiplicativity at coprime integers and the prime-power recurrenceMultiply its generating series by the corresponding quadratic to obtainUnique prime factorization and multiplicativity giveFor a normalized eigenform $c(1)=1$, the prefactor disappears. It must be retained for an arbitrary scalar multiple of that eigenform.

This product identity holds analytically in an absolute-convergence half-plane. The invariant norm of a positive-weight cusp form is bounded, as proved in Question 6. For a period-one Fourier expansion of a modular form,Choosing $y=1/n$ gives $c(n)=O(n^{w/2})$, so $\mathrm{Re}s>w/2+1$ suffices. Absolute convergence then justifies expansion and rearrangement of the Euler product of a Hecke eigenform; no sharper coefficient estimate is needed here.