[proofplan] The proof converts the multiplicative identity for Hecke operators into an identity for their eigenvalues. Since $f$ is a normalized Hecke eigenform, the eigenvalue of $T_n$ on $f$ is exactly the Fourier coefficient $a_n$. Applying the Hecke operator product formula to $f$ therefore gives the divisor-sum relation, and the prime-power recursion is the special case $m=p$ and $n=p^r$. [/proofplan] [step:Identify the Hecke eigenvalues with Fourier coefficients] For each $n \in \mathbb{N}$, let $\lambda_n \in \mathbb{C}$ denote the eigenvalue of $T_n$ on $f$, so that \begin{align*} T_n f = \lambda_n f. \end{align*} Because $f$ is normalized, $a_1 = 1$. By the standard Fourier-coefficient formula for the level-one Hecke operator $T_n$ (citing a result not yet in the wiki: Fourier coefficient formula for Hecke operators), the coefficient of $q^1$ in $T_n f$ is $a_n$. On the other hand, the coefficient of $q^1$ in $\lambda_n f$ is $\lambda_n a_1 = \lambda_n$. Comparing the coefficient of $q^1$ in the equality $T_n f = \lambda_n f$ gives \begin{align*} \lambda_n = a_n. \end{align*} [/step] [step:Apply the Hecke operator product formula to the eigenform] Fix $m,n \in \mathbb{N}$. The level-one Hecke operators satisfy the product identity \begin{align*} T_m T_n = \sum_{d \mid \gcd(m,n)} d^{k-1} T_{mn/d^2} \end{align*} on $S_k(SL_2(\mathbb{Z}))$ (citing a result not yet in the wiki: Hecke operator multiplication formula). Applying both sides to $f$ gives \begin{align*} T_m T_n f = \sum_{d \mid \gcd(m,n)} d^{k-1} T_{mn/d^2} f. \end{align*} Using $T_j f = \lambda_j f$ for each $j \in \mathbb{N}$, the left-hand side becomes \begin{align*} T_m T_n f = T_m(\lambda_n f) = \lambda_n T_m f = \lambda_n \lambda_m f, \end{align*} and the right-hand side becomes \begin{align*} \sum_{d \mid \gcd(m,n)} d^{k-1} T_{mn/d^2} f = \sum_{d \mid \gcd(m,n)} d^{k-1} \lambda_{mn/d^2} f. \end{align*} Therefore \begin{align*} \lambda_m \lambda_n f = \left(\sum_{d \mid \gcd(m,n)} d^{k-1} \lambda_{mn/d^2}\right) f. \end{align*} Since $f \neq 0$, the scalar coefficients are equal: \begin{align*} \lambda_m \lambda_n = \sum_{d \mid \gcd(m,n)} d^{k-1} \lambda_{mn/d^2}. \end{align*} Substituting $\lambda_j = a_j$ for every $j \in \mathbb{N}$ yields \begin{align*} a_m a_n = \sum_{d \mid \gcd(m,n)} d^{k-1} a_{mn/d^2}. \end{align*} [/step] [step:Specialize the divisor formula to prime powers] Let $p$ be prime and let $r \in \mathbb{N}$. Apply the coefficient identity with $m=p$ and $n=p^r$. Since \begin{align*} \gcd(p,p^r) = p, \end{align*} the positive divisors of $\gcd(p,p^r)$ are exactly $1$ 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}} \\ &= a_{p^{r+1}} + p^{k-1} a_{p^{r-1}}. \end{align*} Rearranging gives \begin{align*} a_{p^{r+1}} = a_p a_{p^r} - p^{k-1} a_{p^{r-1}}. \end{align*} This proves both the general Hecke coefficient relation and the stated prime-power recursion. [/step]