Past exam of the probability, statistics and options I course of the University of Oxford / 2010 / A / 1 / 3 / Solution

Reduce the monic integer polynomial modulo $p$. Every $u\in {\mathbb{F}}_p$ satisfies $u^p=u$, so $u^p-u-1=-1\ne 0$. Thus the reduction has no root in ${\mathbb{F}}_p$. If it had a proper factor, part 2 would force such a root, a contradiction. Hence it is irreducible over ${\mathbb{F}}_p$.

The criterion of irreducibility by degree-preserving reduction modulo a prime, which follows from Gauss's lemma, now applies because the polynomial is monic and its degree is preserved on reduction. Consequently $X^p-X-1$ is irreducible in $\mathbb{Q}[X]$.