jmc

We can factor $x^{2017}-2x+1=0$ by writing it as $$\begin{array}{rl}{x}^{2017}-1-2x+2& =(x^{2017}-1)-2(x-1)\\ & =(x-1)(x^{2016}+x^{2015}+\cdots +x+1)-2(x-1)\\ & =(x-1)(x^{2016}+x^{2015}+\cdots +x-1).\end{array}$$Since $x\ne 1,$ we must have $x^{2016}+x^{2015}+\cdots +x-1=0,$ so $x^{2016}+x^{2015}+\cdots +x+1=\boxed{2}.$