Define the arithmetical function $\tau_k $by $$\sum_{n=1}^{\infty} \frac{\tau_k(n)}{n^s}= \zeta(s)^k$$
Obtain a number theoretic interpretation for $ \tau_k(n) $ ie, it is the number of $...$ what?
I realize i want something like $$\sum_{n=1}^{\infty} (\frac{1}{n^s})^k= \zeta(s)^k$$ or something similar in the prime form but i have no idea what this is? Please help.
By the algebra of Dirichlet's series $$ \zeta(s)^k = \sum_{n\geq 1}\frac{1*1*\ldots*1(n)}{n^s} $$ where $1(n)$ is the function which constantly equals $1$, $$ (1*1)(n) = \sum_{d\mid n} 1 = d(n)=\tau(n)=\prod_{p^k\parallel n}(k+1), $$ $$ (1*1*1)(n) = \sum_{d\mid n} \tau(d) = \prod_{p^k\parallel n}\sum_{d\mid p^k}\tau(d)=\prod_{p^k\parallel n}\sum_{j=0}^{k}(j+1)=\prod_{p^k\parallel n}\frac{(k+1)(k+2)}{2}=\frac{\tau(n)\,\tau(n\,\text{rad}(n))}{2^{\omega(n)}} $$ $$ (1*1*1*1)(n) = \prod_{p^k\parallel n}\frac{(k+1)(k+2)(k+3)}{6}=\frac{\tau(n)\,\tau(n\,\text{rad} n)\,\tau(n\,\text{rad}^2 n)}{6^{\omega(n)}} $$
and so on. We exploited the fact that the Dirichlet convolution is associative and the fact that if $f$ is a multiplicative function, $(1*f)(n)=\sum_{d\mid n}f(d)$ is also a multiplicative function.