= Solution
For a nonzero <Hecke eigenform> that is a <cusp form>, $c(1)\ne0$: 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 recurrence
$$
\lambda(p^{r+1})=\lambda(p)\lambda(p^r)-p^{w-1}\lambda(p^{r-1})\quad(r\geq1).
$$
Multiply its generating <series> by the corresponding quadratic to obtain
$$
\sum_{r\geq0}\lambda(p^r)X^r
=\frac1{1-\lambda(p)X+p^{w-1}X^2}.
$$
Unique <prime factorization> and multiplicativity give
$$
\boxed{L(f,s)=\sum_{n\geq1}\frac{c(n)}{n^s}
=c(1)\prod_p\left(1-\lambda(p)p^{-s}+p^{w-1-2s}\right)^{-1}.}
$$
For 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>,
$$
|c(n)|\leq e^{2\pi ny}\int_0^1|f(x+iy)|\,dx\leq C e^{2\pi ny}y^{-w/2}.
$$
Choosing $y=1/n$ gives $c(n)=O(n^{w/2})$, so $\operatorname{Re}s>w/2+1$ suffices. <Absolute convergence> then justifies expansion and rearrangement of the <Euler product of a Hecke eigenform>; no sharper coefficient estimate is needed here.
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