Riemann zeta function.Part 2

This one has a short solution

$\displaystyle\sum _{ n=2 }^{ \infty }{ \left( \zeta (n)-1 \right) } =$

$\displaystyle\sum _{ n=2 }^{ \infty }{ \sum _{ k=2 }^{ \infty }{ \frac { 1 }{ { k }^{ n } } } } =$

$\displaystyle\sum _{ k=2 }^{ \infty }{ \sum _{ n=2 }^{ \infty }{ \frac { 1 }{ { k }^{ n } } } } =$

$\displaystyle\sum _{ k=2 }^{ \infty }{ \frac { 1 }{ k(k-1) } } =$

$1$