jmc

Note that $(x^2+1)(x+1)$ is a factor of $(x^2+1)(x+1)(x-1)=x^4-1.$ Since $$x^{1000}-1=(x^4-1)(x^{996}+x^{992}+x^{988}+\cdots +x^8+x^4+1),$$the remainder when $x^{1000}$ is divided by $(x^2+1)(x+1)$ is $\boxed{1}.$