<s>[INST] Factoring $x^{7} - 1$ into irreducibles over $\mathbb{F}_{2}[x]$ I know this breaks down into $(x - 1)(x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x +1)$, so the task is to show whether the second factor is irreducible over $\mathbb{F}_{2}[x]$. It's quick check that there are no roots, so one way might be to look at all remaining partitions of 6, then divide through by all polynomials of degree 4, 3 and 2. But there must be something faster.
 [/INST] Factoring $x^{7} - 1$ is equivalent to factoring $x^{8} - x$.
Now,  $x^{8} - x=0$ is the equation that defines the finite field with $8$ elements.
Therefore, $x^{8} - x$ has at least one irreducible factor of degree $3$ (here we use that $8=2^3$).
This answer gives a simple argument that $x^3 + x^2 + 1$ and $x^3 + x + 1$ are the only irreducible polynomials of degree $3$ over $\mathbb{F}_{2}$. By the uniqueness of finite fields, both polynomials must divide $x^{8} - x$. Indeed, 
we can easily check that $x^{8} - x=x (x + 1) (x^3 + x + 1) (x^3 + x^2 + 1)$ and so
$x^{7} - 1=(x + 1) (x^3 + x + 1) (x^3 + x^2 + 1)$.
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