To prove $$\sum_{k=0}^{2016}\left(\zeta_k\prod_{~~~~~~j\neq k,\\ 0\leq j\leq 2016}(2017-\zeta_j)\right)=2017,$$ where $\zeta_0,\zeta_1,\cdots,\zeta_{2016}$ are the $2017$-th roots of the unity.
Alternatively, to show $$\sum_{k=0}^{2016}\frac{1}{2017\zeta_{k}-1}=\frac{2017}{2017^{2017}-1}$$ It can be proven by residue theorem. However, is there an elementary proof?
For the first formulation, consider $$f(x):=x^n-1=\prod_{k=0}^{n-1}(x-\zeta_k).$$ We have $$nx^{n-1}=f'(x)=\sum_{k=0}^{n-1}\prod_{j\neq k}(x-\zeta_j),$$ So $$n=xf'(x)-nf(x)=\sum_{k=0}^{n-1}\left(x\prod_{j\neq k}(x-\zeta_j)-\prod_{j=0}^{n-1}(x-\zeta_j)\right)=\sum_{k=0}^{n-1}\zeta_k\prod_{j\neq k}(x-\zeta_j).$$
For the alternative one, note that the roots of $f(x):=x^n-1$ are $\zeta_0,\ldots,\zeta_{n-1}$, so the roots of $$g(x):=n^nf\left(\frac{x+1}{n}\right)=(x+1)^n-n^n=x^n+\ldots+nx+(1-n^n)$$ are $n\zeta_0-1,\ldots,n\zeta_{n-1}-1$. Then by Vieta's formulas $$\sum_{k=0}^{n-1}\frac{1}{n\zeta_k-1}=-\frac{n}{1-n^n}=\frac{n}{n^n-1}.$$ This gives another proof.
