jmc

Note that $$x^{10}+1=(x^2+1)(x^8-x^6+x^4-x^2+1)$$is divisible by $x^8-x^6+x^4-x^2+1,$ and $$x^{100}-1=(x^{10}+1)(x^{90}-x^{80}+x^{70}-x^{60}+x^{50}-x^{40}+x^{30}-x^{20}+x^{10}-1)$$is divisible by $x^{10}+1.$

Hence, $x^{100}-1$ is divisible by $x^8-x^6+x^4-x^2+1,$ so the remainder when $x^{100}$ is divided by $x^8-x^6+x^4-x^2+1$ is $\boxed{1}.$