[proofplan] We apply the [Hecke multiplicativity formula](/theorems/4247) with $m=p$ and $n=p^r$. Since $p$ is prime and $r \geq 1$, the only positive common divisors of $p$ and $p^r$ are $1$ and $p$. Evaluating the two resulting summands gives exactly $T_{p^{r+1}}$ and $p^{k-1}T_{p^{r-1}}$, and then the equivalent recurrence follows by subtracting the second term. [/proofplan] [step:Apply Hecke multiplicativity to the pair $p$ and $p^r$] Let $p$ be a prime number and let $r \geq 1$ be an integer. We apply the Hecke multiplicativity formula to the pair $(m,n) = (p,p^r)$: \begin{align*} T_p T_{p^r} = \sum_{d \mid \gcd(p,p^r)} d^{k-1} T_{pp^r/d^2}. \end{align*} Here the cited input is the standard Hecke multiplicativity formula for normalized Hecke operators; citing a result not yet in the wiki: Hecke multiplicativity formula. [/step] [step:Identify the common divisors as $1$ and $p$] Since $p$ is prime, the positive divisors of $p$ are exactly $1$ and $p$. Since $r \geq 1$, both $1$ and $p$ divide $p^r$. Therefore \begin{align*} \{d \in \mathbb{N} : d \mid \gcd(p,p^r)\} = \{1,p\}. \end{align*} Thus the multiplicativity sum has exactly two terms: \begin{align*} T_p T_{p^r} = 1^{k-1} T_{pp^r/1^2} + p^{k-1} T_{pp^r/p^2}. \end{align*} [/step] [step:Evaluate the two summands] The first summand is \begin{align*} 1^{k-1}T_{pp^r/1^2} = T_{p^{r+1}}. \end{align*} For the second summand, since $pp^r = p^{r+1}$ and $r \geq 1$, division by $p^2$ gives a non-negative exponent: \begin{align*} \frac{pp^r}{p^2} = \frac{p^{r+1}}{p^2} = p^{r-1}. \end{align*} Hence \begin{align*} p^{k-1}T_{pp^r/p^2} = p^{k-1}T_{p^{r-1}}. \end{align*} Substituting these two evaluations into the multiplicativity formula yields \begin{align*} T_p T_{p^r} = T_{p^{r+1}} + p^{k-1}T_{p^{r-1}}. \end{align*} [/step] [step:Rearrange to obtain the recursive form] Subtracting the operator $p^{k-1}T_{p^{r-1}}$ from both sides of the operator identity gives \begin{align*} T_{p^{r+1}} = T_pT_{p^r} - p^{k-1}T_{p^{r-1}}. \end{align*} This is the stated equivalent recurrence, and the proof is complete. [/step]