= Solution
Reduce the monic <integer> <polynomial> modulo the prime $p$. For every $x\in\mathbb F_p$, $x^p=x$, so
$$
x^p-x-1=-1\neq0.
$$
If the reduction were reducible, a proper monic factorization would, by part (2), force a <root> in $\mathbb F_p$, a contradiction. Its reduction is therefore irreducible of degree $p$. The monic reduction criterion, obtained from <Gauss lemma for polynomials> by reducing a hypothetical <integer> factorization, now proves
$$
\boxed{X^p-X-1\text{ is irreducible in }\mathbb Q[X].}
$$
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