There's a simple factorization for the difference of squares:
[tex]a^2-b^2=(a-b)(a+b)[/tex]
Here, we have
[tex]r^8-1=(r^4)^2-1^2=(r^4-1)(r^4+1)[/tex]
But [tex]r^4[/tex] is another square, so we can do more:
[tex]r^4-1=(r^2)^2-1^2=(r^2-1)(r^2+1)[/tex]
Not done yet!
[tex]r^2-1=(r-1)(r+1)[/tex]
Putting everything together, we have
[tex]r^8-1=(r-1)(r+1)(r^2+1)(r^4+1)[/tex]
lemme just add a little bit to the superb reply by @Lammetthash
1 = 1
1² = 1
1³ = 1
1⁸⁹ = 1
1¹⁰⁰⁰⁰⁰⁰⁰⁰⁰ = 1
therefore
[tex]\bf r^8-1\implies r^8-1^8\implies r^{4\cdot 2}-1^{4\cdot 2}\implies (r^4)^2-(1^4)^2 \\\\\\ \begin{array}{rrrr} (r^4-1^4)&&(r^4+1^4)\\[1em] [(r^2)^2-(1^2)^2]\\\\ (r^2-1^2)&(r^2+1^2)\\[1em] (r-1)(r+1)\\ \cline{1-3} \end{array}\\\\(r-1)(r+1)(r^2+1)(r^4+1)[/tex]
