Zeta function identity

Question

How does one prove the zeta function identity

$$\sum_{s=2}^{\infty}\left(1-\sum_{n=1}^{\infty}\frac{1}{n^s}\right)=-1 \;?$$

Answer 1

$$\sum_{k=2}^\infty \left(1-\sum_{n=1}^\infty \frac{1}{n^k}\right) =-\sum_{k=2}^\infty\sum_{n=2}^\infty \frac{1}{n^k} =-\sum_{n=2}^\infty \sum_{k=2}^\infty \frac{1}{n^k} $$

$$ =-\sum_{n=2}^\infty \frac{1/n^2}{1-1/n} =-\sum_{n=2}^\infty\left(\frac{1}{n-1}-\frac{1}{n}\right)=-1. $$