We list prime factor presentations of the binomials $x^n\!-\!1$ in $\mathbb{Q}$ , i.e. in the polynomial ring $\mathbb{Q}[x]$ . The prime factors can always be chosen to be with integer coefficients and the number of the prime factors equals to $\tau(n)$ ; see the proof.

$x-1$

$x^2\!-\!1 = (x+1)(x-1)$

$x^3\!-\!1 = (x^2+x+1)(x-1)$

$x^4\!-\!1 = (x^2+1)(x+1)(x-1)$

$x^5\!-\!1 = (x^4+x^3+x^2+x+1)(x-1)$

$x^6\!-\!1 = (x^2+x+1)(x^2-x+1)(x+1)(x-1)$

$x^7\!-\!1 = (x^6+x^5+x^4+x^3+x^2+x+1)(x-1)$

$x^8\!-\!1 = (x^4+1)(x^2+1)(x+1)(x-1)$

$x^9\!-\!1 = (x^6+x^3+1)(x^2+x+1)(x-1)$

$x^{10}\!-\!1 = (x^4+x^3+x^2+1)(x^4-x^3+x^2-x+1)(x+1)(x-1)$

$x^{11}\!-\!1 = (x^{10}\!+\!x^9\!+\!x^8\!+\!x^7\!+\!x^6\!+\!x^5\!+\!x^4\!+\!x^3\!+\!x^2\!+\!x\!+\!1)(x-1)$

$x^{12}\!-\!1 = (x^4-x^2+1)(x^2+x+1)(x^2-x+1)(x^2+1)(x+1)(x-1)$

$x^{13}\!-\!1 =(x^{12}\!+\!x^{11}\!+\!x^{10}\!+\!x^9\!+\!x^8\!+\!x^7\!+ \!x^6\!+\!x^5\!+\!x^4\!+\!x^3\!+\!x^2\!+\!x\!+\!1)(x\!-\!1)$

$x^{14}\!-\!1 = (x^6+x^5+x^4+x^3+x^2+x+1)(x^6-x^5+x^4-x^3+x^2-x+1)(x+1)(x-1)$

$x^{15}\!-\!1 = (x^8-x^7+x^5-x^4+x^3-x+1)(x^4+x^3+x^2+x+1)(x^2+x+1)(x-1)$

$x^{16}\!-\!1 = (x^8+1)(x^4+1)(x^2+1)(x+1)(x-1)$

$x^{17}\!-\!1 = (x^{16}\!+\!x^{15}\!+\!x^{14}\!+\ldots+\!x^2\!+\!x\!+\!1)(x\!-\!1)$

$x^{18}\!-\!1 = (x^6+x^3+1)(x^6-x^3+1)(x^2+x+1)(x^2-x+1)(x+1)(x-1)$

$x^{19}\!-\!1 = (x^{18}\!+\!x^{17}\!+\!x^{16}\!+\ldots+\!x^2\!+\!x\!+\!1)(x-1)$

$x^{20}\!-\!1 = (x^8-x^6+x^4-x^2+1)(x^4+x^3+x^2+x+1)(x^4-x^3+x^2-x+1)(x^2+1)(x+1)(x-1)$

$x^{21}\!-\!1 = (x^{12}\!-\!x^{11}\!+\!x^9\!-\!x^8\!+\!x^6\!-\!x^4\!+\!x^3\!-\!x\!+\!1)(x^6\!+\!x^5\!+\!x^4\!+\!x^3\!+\!x^2\!+\!x\!+\!1)(x^2\!+\!x\!+\!1) (x\!-\!1)$

$x^{22}\!-\!1 = (x^{10}\!+\!x^9\!+\!x^8\!+\!x^7\!+\!x^6\!+\!x^5\!+\!x^4\!+\!x^3\!+\!x^2\!+\!x\!+\!1)(x^{10}\!-\!x^9\!+\!x^8\!-\!x^7\!+\!x^6\!-\!x^5\!+\!x^4\!-\!x^3\!+\!x^2\!-\!x\!+\!1)(x\!+\!1)(x\!-\!1)$

$x^{23}\!-\!1 = (x^{22}\!+\!x^{21}\!+\!x^{20}\!+\ldots+\!x^2\!+\!x\!+\!1)(x\!-\!1)$

$x^{24}\!-\!1 = (x^8\!-\!x^4\!+\!1)(x^4\!-\!x^2\!+\!1)(x^4\!+\!1)(x^2\!+\!x\!+\!1)(x^2\!-\!x\!+\!1)(x^2\!+\!1)(x\!+\!1)(x\!-\!1)$

Note 1. All factors shown above are irreducible polynomials (in the field $\mathbb{Q}$ of their own coefficients), but of course they (except $x\!\pm\!1$ ) may be split into factors of positive degree in certain extension fields; so e.g. $$x^4\!+\!1 = (x^2\!+\!x\sqrt{2}\!+\!1)(x^2\!-\!x\sqrt{2}\!+\!1)\quad \mathrm{in\,the\,field}\,\,\,\mathbb{Q}(\sqrt{2}).$$ Note 2. The 24 examples of factorizations are true also in the fields of characteristic $\neq 0$ , but then many of the factors can be simplified or factored onwards (e.g. $x^2\!+\!1 \equiv (x\!+\!1)^2$ if the characteristic is 2).