The following expression for the $\frac{1}{\zeta(s)}$ involving the Möbius function is well known:
$$\frac{1}{\zeta(s)}=\sum _{n=1}^{\infty }\frac {\mu(n)}{n^s} \qquad s \in \mathbb{C},\Re(s) > 1$$
There also exists this (recursive) mobius inversion formula for $\zeta(s)$:
\begin{align} \zeta(s) &= 1+\sum _{k=1}^{\infty } {\frac {{\mu} \left( k \right) }{k}\sum _{m=1}^{\infty }{\frac {\zeta \left( kms \right) -1}{m}}}\qquad s \in \mathbb{C},\Re(s) >0, s \ne \frac{1}{i}, i \in \mathbb{N}\\ \text{where the $d$-th derivative is:}\\ \zeta^{d}(s) &= \sum _{k=1}^{\infty } {{\mu} \left( k \right) \sum _{m=1}^{\infty }{\zeta^{d}\left( kms \right)}} (km)^{d-1} \qquad d \in \mathbb{N}, d > 0\\ \text{which is simplest for $d=1$:}\\ \zeta^{1}(s) &= \sum _{k=1}^{\infty } {{\mu} \left( k \right) \sum _{m=1}^{\infty }{\zeta^{1}\left( kms \right)}}\\ \end{align}
Observe that in all equations, $k=m=1$ already yields the LHS, hence the remaining series must always sum to $0$.
Questions:
Could any of these expressions be simplified further (I have tried swapping the sums, which worked fine, however didn't help simplifying things further)?
Taking the anti-derivates for $d$, always correctly recreates the $d-1$ derivative except for the step from $d=1$ to $d=0$ (i.e. when the constants $1$ and $-1$ have to re-appear). Is there a reason for why that final step to $d=0$ is more complicated?