I read that it was known to Euler that $$\sum \frac{\zeta(n) -1}{n} = 1- \gamma$$, but I cannot find a proof of this and am curious how he obtained it. I first tried messing with the definition of $\gamma$, given as the limiting difference between the partial sums of the harmonic series and the natural log, but could not connect it to the zeta function. From whichever method is used to obtain the result, could we likewise obtain $\sum \frac{\zeta(2n) -1}{2n}$ or $\sum \frac{\zeta(2n+1) -1}{2n+1}$?
Could we somehow begin with the result that $\sum (\zeta(n) -1) = 1$?