In general, there is a problem: "Prove that the polynomial $f(x) = x^{p} - x - 1$ is irreducible over $\mathbb{Z}p$ ($p$ is a prime number)". I had an idea to solve this problem using the Butler criterion, which states: "The polynomial $f(x)\in\mathbb{Z}p$ is irreducible if and only if $gcd(f(x), f'(x)) = 1$ and the equation $h^{q} - h = 0$ has in the factor ring $\mathbb{R}=\mathbb{Z}p\backslash(f(x))$ exactly $q$ solutions".
$h = [h(x)]_{f(x)}$, where $h(x) = c_0 + c_1x + ... + c_{n-1}x^{n-1}, n = degf(x)$
The equality $h^{q} - h = 0$ in the ring $\mathbb{R}$ is equivalent to the comparison $h(x)^{q}\equiv h(x)\bmod f(x)$, which is converted to the form $0\cdot c_0+(x^{q} - x)c_1 + (x^{2q} - x^2)c_2+ ... + (x^{(n-1)q} - x^{n-1}) \equiv 0 \bmod f(x)(1)$, since in the ring $\mathbb{Z}p$ the identity $(c_0 +c_1x + ... + c_{n-1}x^{n-1})^{q} = c_0 + c_1x^{q} + ... + c_{n-1}x^{(n-1)q}$
It is clear that $gcd(f(x), f'(x)) = 1$. Next we have $x^{jp}\equiv(x+1)^{j} = x^j + jx^{j-1} + ... \bmod f(x)$ for any $j = 0, 1, ..., p - 1$. But I don't understand what to do next. There are suspicions that it is necessary to somehow present a comparison (1) and find $c_{p-1}, ..., c_1$. But how can we prove irreducibility in the end? Can anyone help?
