$x^4-x-1$ has Galois group $S_4$ over $\mathbb{Q}$

Modulo $2$ the polynomial is $x^4+x+1$. It has no root in ${\mathbb{F}}_2$, and the only irreducible quadratic $x^2+x+1$ does not divide it, so the reduction is irreducible. The reduction test makes $x^4-x-1$ irreducible over $\mathbb{Q}$.

The resolvent formula gives $R(y)=y^3+4y-1$. Its only possible rational roots are $1$ and $-1$, neither of which is a root, so the resolvent cubic is irreducible.

The discriminant is the resolvent discriminant $-4(4{)}^3-27(-1{)}^2=-283$, a negative nonsquare in $\mathbb{Q}$ and in particular nonzero.

The derivative of $x^4-x-1$ is $4x^3-1$, a nonzero element of $\mathbb{Q}[x]$, and $\mathbb{Q}$ has characteristic zero, hence not two.

Steps 1.1 and 1.4 with [L2] make $x^4-x-1$ separable. The hypotheses of [L1] are then supplied by steps 1.1, 1.2, 1.3, and 1.4, so the Galois group is $S_4$.