$x^{p^{2}}-p^{p}$ is irreducible in $\mathbb{Q} [x]$

Question

Let $p$ be a prime number in $\mathbb{N}$.

How to show that $x^{p^{2}}-p^{p}$ is irreducible in $\mathbb{Q} [x]$ ?

Answer 1

We have $$ (x^p)^p-p^p=(x^p-p)((x^p)^{p-1}+p(x^p)^{p-2}+\cdots+p^{p-2}(x^p)+p^{p-1}) $$ meaning the polynomial is reducible.

Answer 2

For $p=2$: $$x^{2^2}-2^2=x^4-4=(x^2+2)(x^2-2)$$

For $p=3$: $$x^{3^2}-3^3=x^9-27=(x^3-3)(x^6+3x^3+9)$$

For $p=5$: $$x^{5^2}-5^5=x^{25}-3125=(x^5-5)(x^{20}+5x^{15}+25x^{10}+125x^{5}+625)$$

In fact, $x^{p^2}-p^p=(x^p-p)\sum_{i=0}^{p-1}(p^ix^{p^2-(i+1)p})$

The polynomial is always factorizable into two smaller polynomials with integer coefficients. However, these smaller polynomials have no rational roots.