I am having difficulty determining the number of irreducible factors of $x^{127} - 1$ over $\mathbb{Z}_2$.
I know that all irreducible factors of $1$ and $7$ are divisors of $x^{127} - 1$, since $x^{127} = x^{2^7 - 1}$, and $1, 7$ are the divisors of $7$. This is derived from the theorem that $x^{(p^n)}−x$ is the product of all the irreducible polynomials of degree $d,d \mid n$ in $\mathbb{F}_p[x]$
I know that there are of course 2 irreducible polynomials of degree $1$ over $\mathbb{Z}_2$, which are $x$ and $x + 1$, but I'm unsure about how to determine how many irreducible polynomials of degree $7$ there is (is there an efficient and simple algorithm for doing so)? Furthermore, even I have all the irreducible polynomials of degree $7$, performing long division to find the other divisors of $x^{127} - 1$ is unfeasible to do by hand. Are there any tricks I can use to solve tis problem.