I want to prove that
$p(x):=x^n-x-1 \in \mathbb Q[x]$ for $n\ge 2$ is irreducible.
My attempt.
GCD of coefficients is $1$, $\mathbb Q$ is the field of fractions of $\mathbb Z$, and $\mathbb Z$ is UFD. Hence, $p(x)$ is irreducible over $\mathbb Q$ iff it's irreducible over $\mathbb Z$ (Gauss's lemma).
Let $m\in \mathbb Z$ such that $\varphi(m)=n$ (Euler's totient). Make reduction of $p(x)$ by modulo $m$. Because of $\overline{x^n}=\overline{x^{\varphi(m)}}=\overline{1}$, we get $\overline{p(x)}=\overline{1-x-1}=\overline{-x}$, which is irreducible. Hence, $p(x)$ is irreducible.
Does this proof is correct?
UPDATE. Thanks to Calvin Lin. My mistake is: not for all $n$ we can find such $m$. OK, but as for the rest, does my proof is correct for such $n$, that $n=\varphi(m)$ for an integer $m$? And can it be some changed for all $n$, i.e. can we find such modulo that $\overline{p(x)}$ is irreducible for every $n\ge 2$?
I doubt an easy proof of the irreducibility exists in general. If $n$ is a prime, then the polynomial is Artin-Schreier and handled easily.
Selmer gave a clever proof in the general case, working explicitly with the roots of the polynomial in $\mathbb{C}$. See E. S. Selmer, On the irreducibility of certain trinomials, Math. Scand. 4 (1956), 287-302, available here.