[proofplan] The proof uses the normalized Hecke product formula assumed in the statement, which expresses $T_mT_n$ as an explicit finite sum over the common divisors of $m$ and $n$ with coefficients $d^{k-1}$. That expression is symmetric in $m$ and $n$, so the same formula gives $T_nT_m$. Once the generating Hecke operators commute pairwise, every element of the algebra they generate commutes with every other element. [/proofplan] [step:Apply the normalized Hecke product formula to two generators] Fix $m,n \in \mathbb{N}$. Let $\gcd(m,n)$ denote the greatest common divisor of $m$ and $n$. The operators $(T_r)_{r \in \mathbb{N}}$ are the normalized Hecke operators on $V_k$, so the product formula in the theorem statement applies to $T_m$ and $T_n$ and gives \begin{align*} T_mT_n = \sum_{d \mid \gcd(m,n)} d^{k-1} T_{mn/d^2}. \end{align*} Here the sum is finite because the set of positive divisors of $\gcd(m,n)$ is finite, and for every divisor $d \mid \gcd(m,n)$ the integer $mn/d^2$ is positive. [/step] [step:Use symmetry of the divisor formula to reverse the order] Since $\gcd(m,n)=\gcd(n,m)$ and $mn = nm$, the indexing set and every summand in the preceding formula are unchanged when $m$ and $n$ are interchanged. Applying the same normalized Hecke product formula from the theorem statement to $T_nT_m$ gives \begin{align*} T_nT_m = \sum_{d \mid \gcd(n,m)} d^{k-1} T_{nm/d^2} = \sum_{d \mid \gcd(m,n)} d^{k-1} T_{mn/d^2} = T_mT_n. \end{align*} Thus $T_mT_n=T_nT_m$ for all $m,n \in \mathbb{N}$. [/step] [step:Extend pairwise commutativity from generators to the whole Hecke algebra] By definition, $\mathbb{T}_k$ is generated by the family $\{T_n : n \in \mathbb{N}\}$ as a subalgebra of $\operatorname{End}_{\mathbb{C}}(V_k)$. Since every pair of generators commutes, any finite product of generators may be reordered without changing the resulting endomorphism. Therefore any two finite $\mathbb{C}$-linear combinations of such products commute. Hence every two elements of $\mathbb{T}_k$ commute, so $\mathbb{T}_k$ is commutative. [/step]