Factor x^n - 1 for n = 6, 7, 8,..., 22.
Do I need to show the answer for each n? What is the proper way to show this?
But notice that if the exponent is not prime then some further factorisation may be possible. For example, $\displaystyle x^6-1 = (x^3)^2-1 = (x^3+1)(x^3-1) = (x+1)(x^2-x+1)(x-1)(x^2+x+1)$.
[I'm assuming that you are looking for factorisations where the coefficients are rational numbers. If you're working over the complex numbers then you can factorise $\displaystyle x^n-1$ into n linear factors. If you are doing it over the real numbers then you can get a factorisation into linear and quadratic factors (but that is just as hard as doing it over the complex numbers).]