Initially, I thought that calculating$$\int_\gamma \frac{\zeta(z)}{(z-1)^n}dz$$ directly, where $n \in \mathbb{Z}$ and $\gamma$ is an anticlockwise contour around $z=1$ with winding number $1$, would be a nice way of doing this, but I think this is impossible. Is there a nice way of showing that $$\zeta(s)=\frac{1}{s-1}+\sum_{n \ge 1}\frac{(-1)^n}{n!}\lim_{m \to \infty}\sum_{j=1}^m\frac{\log j^n}{j}- \frac{\log j^{n+1}}{n_1}?$$
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I added this answer recently. The Laurent Series for $\zeta(s)$ at $s=1$ is derived. Unfortunately, no contour integration is used. – robjohn Feb 24 '16 at 03:41