This is the statement I am trying to prove or disprove:

Let $\displaystyle p(x) = \frac{x^m - 1}{x - 1}$ where $\displaystyle m = p^n$ for some prime number p and some positive integer n. Then $\displaystyle p(x)$ has exactly n irreducible factors over the integers.

I can see how such polynomials could be factored into n factors but I am having difficulty proving or disproving that the factors are irreducible.

For instance, if p is 2 and n is 4 we have:

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

Clearly these factors are irreducible.

If p is 3 and n is 3 we have:

$\displaystyle p(x) = \frac{x^{27} - 1}{x - 1} = (x^{18} + x^9 + 1)(x^6 + x^3 + 1)(x^2 + x + 1)$

If p is 5 and n is 3 we have:

$\displaystyle p(x) = \frac{x^{125} - 1}{x - 1} = (x^{100} + x^{75} + x^{50} + x^{25} + 1)(x^{20} + x^{15} + x^{10} + x^5 + 1)(x^4 + x^3 + x^2 + x + 1)$

So I see the pattern that produces these factors, but I don't know how to show that all of them are irreducible if they are.