Summing Zeta

$\begin{aligned}\sum_{n=2}^\infty\left(\zeta(n)-1\right)=&\sum_{n=2}^\infty\left[\left(1+\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{4^n}+\dots\right)-1\right]\\ =&\sum_{n=2}^\infty\left(\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{4^n}+\dots\right)\\ =&\quad\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\dots\\ &+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\dots\\ &+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\dots\\ &\quad\vdots\qquad\vdots\qquad\vdots\\ &\text{Grouping in columns instead of rows:}\\ =&\sum_{n=2}^\infty\left(\frac{1}{n^2}+\frac{1}{n^3}+\frac{1}{n^4}+\dots\right)\\ &\text{Using the formula for the sum of an infinite geometric series:}\\ =&\sum_{n=2}^\infty\left(\frac{\frac{1}{n^2}}{1-\frac{1}{n}}\right)\\ =&\sum_{n=2}^\infty\left(\frac{1}{n(n-1)}\right)\\ =&\lim_{N\to\infty}\sum_{n=2}^N\left(\frac{1}{n-1}-\frac{1}{n}\right)\\ =&\lim_{N\to\infty}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{N-2}-\frac{1}{N-1}+\frac{1}{N-1}-\frac{1}{N}\right)\\ =&\lim_{N\to\infty}\left(1-\frac{1}{N}\right)\\ =&1\end{aligned}$

Similar solution with @Joseph Newton 's

$\begin{aligned} S & = \sum_{n=2}^\infty (\zeta (n) - 1) \\ & = \sum_{n=2}^\infty \left(\sum_{\color{#3D99F6}k=1}^\infty \frac 1{k^n} - 1\right) \\ & = \sum_{n=2}^\infty \sum_{\color{#D61F06}k=2}^\infty \frac 1{k^n} \\ & = \sum_{\color{#D61F06}k=2}^\infty \sum_{n=2}^\infty \frac 1{k^n} \\ & = \sum_{k=2}^\infty \frac 1{k^2}\left(\frac 1{1-\frac 1k}\right) \\ & = \sum_{k=2}^\infty \frac 1{k(k-1)} \\ & = \sum_{k=2}^\infty \left(\frac 1{k-1} - \frac 1k\right) \\ & = \boxed{1} \end{aligned}$