[proofplan] The proof uses the Hecke relations for [Fourier coefficients of a normalized Hecke eigenform](/theorems/4251). These relations first give multiplicativity away from common prime factors, so absolute convergence allows the Dirichlet series to factor into local prime-[power series](/page/Power%20Series). For each prime $p$, the same Hecke relation gives a second-order recurrence for the coefficients $a_{p^r}$, and this recurrence identifies the local generating function with the reciprocal of $1-a_pX+p^{k-1}X^2$. Substituting $X=p^{-s}$ and multiplying the local identities gives the Euler product. [/proofplan] [step:Extract multiplicativity and the prime-power recurrence from the Hecke relations] Let $\mathbb{N}=\{1,2,3,\dots\}$ denote the set of positive integers. We use the standard Hecke relation for Fourier coefficients of normalized Hecke eigenforms: for all $m,n \in \mathbb{N}$, \begin{align*} a_m a_n=\sum_{d \mid \gcd(m,n)} d^{k-1} a_{mn/d^2}. \end{align*} This external Hecke relation is the only input about Hecke operators used in the proof. If $\gcd(m,n)=1$, the only positive divisor of $\gcd(m,n)$ is $1$, and therefore \begin{align*} a_m a_n=a_{mn}. \end{align*} Thus the coefficient sequence $(a_n)_{n \geq 1}$ is multiplicative. Now fix a prime number $p$ and an integer $r \geq 1$. Applying the Hecke relation with $m=p$ and $n=p^r$ gives \begin{align*} a_p a_{p^r} &=\sum_{d \mid p} d^{k-1} a_{p^{r+1}/d^2} \\ &=a_{p^{r+1}}+p^{k-1}a_{p^{r-1}}. \end{align*} Rearranging yields the recurrence \begin{align*} a_{p^{r+1}}=a_p a_{p^r}-p^{k-1}a_{p^{r-1}} \end{align*} for every $r \geq 1$. [guided] Let $\mathbb{N}=\{1,2,3,\dots\}$ denote the set of positive integers. The Hecke relation is the algebraic input. Since $f$ is a normalized Hecke eigenform, its Fourier coefficients satisfy, for all positive integers $m$ and $n$, \begin{align*} a_m a_n=\sum_{d \mid \gcd(m,n)} d^{k-1} a_{mn/d^2}. \end{align*} This external Hecke relation is the only input about Hecke operators used in the proof. First suppose $\gcd(m,n)=1$. Then the set of positive divisors of $\gcd(m,n)$ is the singleton $\{1\}$. The Hecke relation reduces to \begin{align*} a_m a_n=1^{k-1}a_{mn}=a_{mn}. \end{align*} This is exactly multiplicativity of the coefficient sequence. Next fix a prime number $p$ and an integer $r \geq 1$. We apply the same Hecke relation with $m=p$ and $n=p^r$. Since $\gcd(p,p^r)=p$, its positive divisors are $1$ and $p$. Hence \begin{align*} a_p a_{p^r} &=\sum_{d \mid p} d^{k-1} a_{p^{r+1}/d^2} \\ &=1^{k-1}a_{p^{r+1}}+p^{k-1}a_{p^{r+1}/p^2} \\ &=a_{p^{r+1}}+p^{k-1}a_{p^{r-1}}. \end{align*} Solving this identity for $a_{p^{r+1}}$ gives \begin{align*} a_{p^{r+1}}=a_p a_{p^r}-p^{k-1}a_{p^{r-1}}. \end{align*} This recurrence is the local relation that will determine the Euler factor at $p$. [/guided] [/step] [step:Factor the absolutely convergent Dirichlet series into prime-power series] Fix $s \in \mathbb{C}$ such that \begin{align*} \sum_{n=1}^{\infty} |a_n n^{-s}|<\infty. \end{align*} Let $\mathcal{P}$ denote the set of prime numbers. For a finite subset $P \subset \mathcal{P}$, define \begin{align*} N(P)=\{n \in \mathbb{N}: \text{every prime divisor of } n \text{ lies in } P\}. \end{align*} By unique factorization and multiplicativity of $(a_n)_{n \geq 1}$, \begin{align*} \prod_{p \in P}\left(\sum_{r=0}^{\infty} a_{p^r}p^{-rs}\right) = \sum_{n \in N(P)} a_n n^{-s}. \end{align*} The series on the right is absolutely convergent because it is a subseries of the absolutely convergent series defining $L(f,s)$. As $P$ increases through finite subsets of $\mathcal{P}$, the sets $N(P)$ increase to all of $\mathbb{N}$. Absolute convergence gives \begin{align*} \lim_{P}\sum_{n \in N(P)} a_n n^{-s} = \sum_{n=1}^{\infty}a_n n^{-s} = L(f,s), \end{align*} where the limit is over the directed set of finite subsets of $\mathcal{P}$ ordered by inclusion. Therefore \begin{align*} L(f,s)=\prod_{p}\left(\sum_{r=0}^{\infty}a_{p^r}p^{-rs}\right). \end{align*} [guided] We now justify the passage from a Dirichlet series to an Euler product. Fix a complex number $s$ in the region of absolute convergence, so \begin{align*} \sum_{n=1}^{\infty} |a_n n^{-s}|<\infty. \end{align*} Let $\mathcal{P}$ be the set of prime numbers. For a finite subset $P \subset \mathcal{P}$, define \begin{align*} N(P)=\{n \in \mathbb{N}: \text{every prime divisor of } n \text{ lies in } P\}. \end{align*} Thus $N(P)$ consists exactly of those positive integers whose prime factorization uses only primes from $P$. For such a finite set $P$, each $n \in N(P)$ has a unique expression \begin{align*} n=\prod_{p \in P} p^{r_p}, \end{align*} where each $r_p$ is a nonnegative integer and all but finitely many exponents are already accounted for because $P$ is finite. Since the coefficients are multiplicative, repeated use of $a_{mn}=a_m a_n$ for coprime $m$ and $n$ gives \begin{align*} a_n=\prod_{p \in P} a_{p^{r_p}}. \end{align*} Also, \begin{align*} n^{-s}=\prod_{p \in P} p^{-r_p s}. \end{align*} Multiplying the finitely many local series therefore gives \begin{align*} \prod_{p \in P}\left(\sum_{r=0}^{\infty} a_{p^r}p^{-rs}\right) = \sum_{(r_p)_{p \in P}} \prod_{p \in P} a_{p^{r_p}}p^{-r_p s} = \sum_{n \in N(P)} a_n n^{-s}. \end{align*} The rearrangement is legitimate because the final series is a subseries of the absolutely convergent series \begin{align*} \sum_{n=1}^{\infty}|a_n n^{-s}|. \end{align*} Finally, as the finite set $P$ grows through all finite subsets of $\mathcal{P}$, the sets $N(P)$ exhaust $\mathbb{N}$. Absolute convergence permits passage to the limit over these finite partial Euler products: \begin{align*} \lim_{P}\sum_{n \in N(P)} a_n n^{-s} = \sum_{n=1}^{\infty}a_n n^{-s} = L(f,s). \end{align*} Hence \begin{align*} L(f,s)=\prod_p\left(\sum_{r=0}^{\infty}a_{p^r}p^{-rs}\right). \end{align*} [/guided] [/step] [step:Compute each local prime-power generating function] Fix a prime number $p$. Define the formal power series $A_p(X) \in \mathbb{C}[[X]]$ by \begin{align*} A_p(X)=\sum_{r=0}^{\infty}a_{p^r}X^r. \end{align*} The identity below is interpreted in the formal power series ring $\mathbb{C}[[X]]$. Since $a_1=1$, we have $a_{p^0}=a_1=1$. Multiplying by $1-a_pX+p^{k-1}X^2$ gives \begin{align*} (1-a_pX+p^{k-1}X^2)\sum_{r=0}^{\infty}a_{p^r}X^r &=a_1+(a_p-a_pa_1)X \\ &\quad+\sum_{r=2}^{\infty}\left(a_{p^r}-a_p a_{p^{r-1}}+p^{k-1}a_{p^{r-2}}\right)X^r. \end{align*} The coefficient of $X$ is zero because $a_1=1$. For $r \geq 2$, the recurrence from the first step gives \begin{align*} a_{p^r}-a_p a_{p^{r-1}}+p^{k-1}a_{p^{r-2}}=0. \end{align*} Therefore \begin{align*} (1-a_pX+p^{k-1}X^2)\sum_{r=0}^{\infty}a_{p^r}X^r=1, \end{align*} and hence \begin{align*} \sum_{r=0}^{\infty}a_{p^r}X^r=\left(1-a_pX+p^{k-1}X^2\right)^{-1}. \end{align*} [guided] We now compute the local factor at a fixed prime $p$. Introduce the prime-power formal power series $A_p(X) \in \mathbb{C}[[X]]$ by \begin{align*} A_p(X)=\sum_{r=0}^{\infty}a_{p^r}X^r. \end{align*} We work in the formal power series ring $\mathbb{C}[[X]]$, so no convergence in $X$ is being asserted at this stage. Since $f$ is normalized, $a_1=1$, and because $p^0=1$, this says \begin{align*} a_{p^0}=1. \end{align*} The recurrence from the Hecke relation has the same shape as the denominator we want. Therefore we multiply $A_p(X)$ by $1-a_pX+p^{k-1}X^2$ and compare coefficients: \begin{align*} (1-a_pX+p^{k-1}X^2)A_p(X) &=\left(\sum_{r=0}^{\infty}a_{p^r}X^r\right) -a_pX\left(\sum_{r=0}^{\infty}a_{p^r}X^r\right) +p^{k-1}X^2\left(\sum_{r=0}^{\infty}a_{p^r}X^r\right) \\ &=a_1+(a_p-a_pa_1)X \\ &\quad+\sum_{r=2}^{\infty}\left(a_{p^r}-a_p a_{p^{r-1}}+p^{k-1}a_{p^{r-2}}\right)X^r. \end{align*} The constant term is $a_1=1$. The coefficient of $X$ is \begin{align*} a_p-a_pa_1=a_p-a_p=0. \end{align*} For every $r \geq 2$, the recurrence with $r-1$ in place of $r$ says \begin{align*} a_{p^r}=a_p a_{p^{r-1}}-p^{k-1}a_{p^{r-2}}. \end{align*} Equivalently, \begin{align*} a_{p^r}-a_p a_{p^{r-1}}+p^{k-1}a_{p^{r-2}}=0. \end{align*} Thus every coefficient except the constant term vanishes, and we obtain \begin{align*} (1-a_pX+p^{k-1}X^2)A_p(X)=1. \end{align*} Solving for $A_p(X)$ gives the local identity \begin{align*} \sum_{r=0}^{\infty}a_{p^r}X^r=\left(1-a_pX+p^{k-1}X^2\right)^{-1}. \end{align*} [/guided] [/step] [step:Substitute $X=p^{-s}$ and assemble the Euler product] For the fixed $s$ in the absolute convergence region and each prime $p$, the prime-power series converges absolutely because \begin{align*} \sum_{r=0}^{\infty}|a_{p^r}p^{-rs}| \end{align*} is a subseries of $\sum_{n=1}^{\infty}|a_n n^{-s}|$. Therefore the formal identity from the previous step may be evaluated at $X=p^{-s}$. This gives, for every prime $p$, \begin{align*} \sum_{r=0}^{\infty}a_{p^r}p^{-rs} = \left(1-a_p p^{-s}+p^{k-1}p^{-2s}\right)^{-1} = \left(1-a_p p^{-s}+p^{k-1-2s}\right)^{-1}. \end{align*} Combining this identity with the prime-power factorization of $L(f,s)$ yields \begin{align*} L(f,s) &=\prod_p\left(\sum_{r=0}^{\infty}a_{p^r}p^{-rs}\right) \\ &=\prod_p\left(1-a_p p^{-s}+p^{k-1-2s}\right)^{-1}. \end{align*} This is the desired Euler product throughout the half-plane of absolute convergence. [/step]