polynomials - Factor $x^8-x$ in $Bbb Z[x]$ and in $Bbb Z_2[x]$

Factor $x^8-x$ in $\Bbb Z[x]$ and in $\Bbb Z_2[x]$

Here what I get is $x^8-x=x(x^7-1)=x(x-1)(1+x+x^2+\cdots+x^6)$ now what next? Help in both the cases in $\Bbb Z[x]$ and in $\Bbb Z_2[x]$

Edit: I think $(1+x+x^2+\cdots+x^6)$ is cyclotomic polynomial for $p=7$ so it is irred over $\Bbb Z$. Now the problem remains for $\Bbb Z_2[x]$

Answer

After my edit I finally got the answer it was under my nose the polynomial $1+\cdots +x^6$ does not have zeros at $0$ and $1$ so it can't be factored as a multiple of a one degree polynomial and one other as a whole.

Edit: But as Lubin mentioned in the comment $1+\cdots +x^6=(1+x+x^3)(1+x^2+x^3)$