This is a fairly simple question but it seems really confusing for me. So basically I am reading the proof of the composition formula of Hecke operators: $$T(m)T(n)=\sum_{d\mid (m,n)}d^{k-1}T(mn/d^2)$$ The proof is similar to what we did about $\tau(m)\tau(n)$. We just need to prove that $$T(p^{r+1})=T(p)T(p^r)-p^{k-1}T(p^{r-1})\text{ or }T(p)T(p^r)=T(p^{r+1})+p^{k-1}T(p^{r-1})$$ It is shown that $$\begin{align*} (T(p^{r})f)(\tau) &=p^{rk-r}\sum_{j=0}^r p^{-jk}\sum_{b=0}^{p^j-1}f\left(\frac{p^r\tau+bp^j}{p^{2j}}\right)\\ &=p^{-r}\sum_{\substack{0\leqslant j\leqslant r\\ 0\leqslant b_j<p^j}}p^{(r-j)k}f\left(\frac{p^{r-j}\tau+b_j}{p^{j}}\right) \end{align*}$$ and $$(T(p)g)(\tau)=p^{k-1}g(p\tau)+\frac{1}{p}\sum_{b=0}^{p-1}g\left(\frac{\tau+b}{p}\right)$$ So we just need to consider the composition $(T(p)T(p^r)f)(\tau)$ In Apostol's book it is said that $$\begin{align*} (T(p)T(p^r)f)(\tau) ={}&p^{k-1-r}\sum_{\substack{0\leqslant j\leqslant r\\ 0\leqslant b_j<p^j}}p^{(r-j)k}f\left(\frac{p^{r-j+1}\tau+pb_j}{p^{j}}\right)\\ &+p^{-1-r}\sum_{\substack{0\leqslant j\leqslant r\\ 0\leqslant b_j<p^j}}\sum_{b=0}^{p-1}f\left(\frac{p^{r-j}\tau+b_j+bp^j}{p^{j+1}}\right) \end{align*}$$ I don't quite understand where that $pb_j$ comes from in the first sum. It should be $g(p\tau)$ where $g=T(p^r)f$ and thus $g(p\tau)=(T(p^{r})f)(p\tau)$. Shouldn't this be something like $$(T(p^{r})f)(p\tau)=p^{-r}\sum_{\substack{0\leqslant j\leqslant r\\ 0\leqslant b_j<p^j}}p^{(r-j)k}f\left(\frac{p^{r-j}\boldsymbol{(p\tau)}+b_j}{p^{j}}\right)?$$
I tried to read some other materials and found no explanation. In Serre's A Course in Arithmetic this is proved by the Fourier expansion of $T_m$. In Gunning's Lectures on Modular Forms, this was presented in the same way without explanation. I believe it's a very minor issue but I just couldn't get it. Any help will be appreciated and thanks in advance.