jmc

Note that $$(x^2+1)(x^8-x^6+x^4-x^2+1)=x^{10}+1.$$Also, $x^{10}+1$ is a factor of $x^{2010}+1$ via the factorization $$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}{b}^2+\cdots +b^{n-1})$$where $n$ is odd, so $x^{10}+1$ is a factor of $x^5(x^{2010}+1)=x^{2015}+x^5.$

So, when $x^{2015}+1=x^{2015}+x^5+(-x^5+1)$ is divided by $x^8-x^6+x^4-x^2+1,$ the remainder is $\boxed{-x^5+1}.$