I'm reposting this question with a clarification and deleting the old one.
From the comments on that question I learned that $$\zeta^k(s)=\sum_{n=1}^\infty\frac{d_k(n)}{n^s}$$ where $d_k(n)$ is the number of ways of writing $n$ as a product of $k$ positive integers (I assume that the order of the terms in the products matters in order to make $d_2(n)=d(n)$ but check me on this).
I had previously said I was interested in the dirichlet series of $\zeta^k(s)$ for $k$ being an integer, but I realized that I need $k$ to be any positive real number.
According to this generating function for the coefficients of such a dirichlet series, $k\in\mathbb{R}$ is valid.
What would the coefficients of our dirichlet series be?
Let $$d_k(n) = \# \{ m \in \mathbb{Z}_{\ge 1}^k,\prod_{i=1}^k m_i = n\}, \qquad e_k(n) = \# \{ m \in \mathbb{Z}_{\ge 2}^k,\prod_{i=1}^k m_i = n\}$$ For $k \in \mathbb{Z}_{\ge 0}$ : $$\zeta(s)^k = (\sum_{m=1}^\infty m^{-s})^k = \sum_{m \in \mathbb{Z}_{\ge 1}^k } \prod_{i=1}^k m_i^{-s} =\sum_{n=1}^\infty n^{-s} d_k(n)$$
$$(\zeta(s)-1)^k = (\sum_{m=2}^\infty m^{-s})^k = \sum_{m \in \mathbb{Z}_{\ge 2}^k } \prod_{i=1}^k m_i^{-s} =\sum_{n=1}^\infty n^{-s} e_k(n)$$
For $a \in \mathbb{C}\setminus - \mathbb{N}$ since $\zeta(s)-1 \to 0$ as $\Re(s) \to \infty$ : $$\zeta(s)^a=(1+(\zeta(s)-1))^a = \sum_{k=0}^\infty {a \choose k}(\zeta(s)-1)^k =\sum_{k=0}^\infty {a \choose k} \sum_{n=1}^\infty n^{-s} e_k(n) \\=\sum_{n=1}^\infty n^{-s}\sum_{k=0}^\infty {a \choose k} e_k(n) $$