Euler product of the $L$-function implies Heck eigenform Suppose $f(z)=\sum c(n)q^n$ and $L(f,s)=\sum c(n)n^{-s}$ have the Euler product 
$$L(f,s)=\prod_{p}\frac{1}{1 −c(p)p^{−s} + p^{k−1−2s}}.$$
I wonder if/want to show that $f$ is actually a Hecke eigenform (the converse of this claim can be found in any book about modular forms). 
My thoughts: It follows from the above assumption that
$$1=\prod_{p} (1 −c(p)p^{−s} + p^{k−1−2s})(\sum_m c(p^{-ms})p^{-ms}).$$
Specifically I have to following two subquestions:
(1)I don't know if there is a "Global-local principle" to show that we can drop the product sign, i.e. for each $p$, we have
$$1=(1 −c(p)p^{−s} + p^{k−1−2s})(\sum_m c(p^{-ms})p^{-ms})$$
(2) If (1) is indeed true, then we have the recursive formula:
$$c(p)c(p^n)=c(p^{n+1})+p^{k-1}c(p^{n-1})$$
I wonder how to use this recursive formula as the character property of $c(n)$ to show that $f$ is a Hecke eigenform.
 A: If $$f(z)=\sum_{n=1}^\infty a_n(f)e^{2i\pi nz}\in S_k(SL_2(\Bbb{Z}))$$ then we define the Hecke operators
$$T_m f\in S_k(SL_2(\Bbb{Z})),\qquad a_n(T_p f) = a_{pn}(f) +1_{p|n}p^{k-1} a_{p/n}(f) , \qquad \gcd(n,m)=1,T_nT_m = T_{nm}, T_{p^r} = T_pT_{p^{r−1}} − p^{k−1}T_{p^{r−2}}$$
If also $$L(f,s) =\sum_{n=1}^\infty a_n(f) n^{-s}= \prod_p \frac1{1-a_p(f) p^{-s}+p^{k-1-2s}}$$ has an Euler product then the $a_n(f)$ are multiplicative so that $$\gcd(n,m)=1, a_m(T_n f)  = a_{nm}(f)=a_n(f)a_m(f)$$ 
For the remaining case
$$ T_p (a_{p^l}(f)) =a_{p{l+1}}(f)+ p^{k-1}a_{p^{l-1}}(f)= a_p(f) a_{p^l}(f), \qquad  T_p (a_{p^l n}(f))=a_n(f) T_p (a_{p^l}(f))$$ thus by induction using the multiplicativity there are some $b_n(f)$ such that for all $n,m, a_m(T_n f) = b_n(f)a_m(f)$ and looking at $\gcd(n,m)=1$ we get $b_n(f)=a_n(f)$ ie.
$$T_n f=  a_n(f)f$$
The case $f\in S_k(\Gamma_0(N))$ is similar : we define the Hecke operators similarly in term of the coefficients, we only change the recursion for $T_{p^r},p | N$ to  $$T_{p^r} = T_pT_{p^{r−1}} − p^{k−1}T_{p^{r−2}} 1_{p \nmid N}$$
and the Euler product for $f$ becomes $$ \prod_p \frac1{1-a_p(f) p^{-s}+1_{p \nmid d}p^{k-1-2s}}$$ thus $f$ is an eigenfunction of all the Hecke operators only for $d=N$.
If $f$ has an Euler product of any form then it is an eigenfunction of the Hecke operators $\gcd(n,N)=1$, this follows from finding a basis of eigenforms.
