In Serre's $A \, Course\, In \,Arithmetic$, it says the following:
$\sum\limits_{n=1}^{\infty}c(n)/n^s= \prod\limits_{p \,\rm prime}\frac{1}{1-c(p)p^{-s}+p^{2k-1-2s}}$
$\Longleftrightarrow$
$\left\{\begin{array}{ll} c(m)c(n)=c(mn) && (m,n)=1 \\ c(p)c(p^n)=c(p^{n+1})+p^{2k-1}c(p^{n-1})&& p \,{\rm prime} \,{\rm and} \, n\geq 1, \end{array}\right.$
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It proves the $\Longleftarrow$ part in the book by $\sum\limits_{n=1}^{\infty}c(n)/n^s=\prod\limits_{p \, {\rm prime}}(\sum\limits_{n=0}^\infty{c(p^n)p^{-ns}})$, and expands $(\sum\limits_{n=0}^\infty{c(p^n)p^{-ns}})(1-c(p)p^{-s}+p^{2k-1-2s})$ to get a power series in $p^{-s}$, and by relations of $c(mn), c(p^n)$ above, get it is equal to $c(1)=1$, hence $\prod\limits_{p \,\rm prime}(\sum\limits_{n=0}^\infty{c(p^n)p^{-ns}})({1-c(p)p^{-s}+p^{2k-1-2s}})=\prod\limits_{p \,\rm prime}1=1$, and the $\Longleftarrow$ part is done.
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But the book doesn't prove the $\Longrightarrow$ part. In order for $\sum\limits_{n=1}^{\infty}c(n)/n^s=\prod\limits_{p \, {\rm prime}}(\sum\limits_{n=0}^\infty{c(p^n)p^{-ns}})$ to be true, we must have $c(mn)=c(m)c(n)$ first, but this is the property to be proved, so how to prove the $\Longrightarrow$ part?
You start with a Dirichlet sum $ \sum_{n\geq 1} \frac{c(n)}{n^s} $ and get a product over primes: this is possible if the coefficients $c(n)$ are given by a multiplicative function (then $c(1)=1$), in other words it's an Euler product. In general you can only say that $$ \sum_{n\geq 1} \frac{c(n)}{n^s} =\prod_p\sum_{j\geq0}\frac{c(p^{js})}{p^{js}}\;. $$ If the function $c$ is completely multiplicative, you get $$ \prod_p\sum_{j\geq0}\frac{c(p^{js})}{p^{js}}= \prod_p\sum_{j\geq0}\frac{c(p^{s})^j}{p^{js}}=\prod_p\frac1{1-c(p)p^{-s}}\;, $$ as for the Dirichlet $L$-series, where $c(n)=\chi(n)$ is a Dirichlet character. In Serre's example, you don't have this expression, so you can deduce that $c$ is only multiplicative but not completely multiplicative. Then you should focus on the expression for a single prime $p$, investigating the powers of such $p$ to prove the relation you want.