Show that $f(x) = x^p -x -1 \in \Bbb{F}_p[x]$ is irreducible over $\Bbb{F}_p$ for every $p$. Let $p$ be a prime.
a) Show that $f$ has no roots in $\Bbb{F}_p$.

Let $F^*$ be the multiplicative group of $\Bbb{F}_p$. Then, by lagrange's thoerem for all nonzero $\alpha \in \Bbb{F}_p$, $\alpha^{p-1} = 1 \implies \alpha^p=\alpha \implies \alpha^p-\alpha=0$. Of course $0^p=0$, so this is true for all elements of $F$ and not just the nonzero ones. But then $\alpha^p - \alpha - 1 = -1$ for all $\alpha \in \Bbb{F}_p$ and so it must have no roots in $\Bbb{F}_p$. I could have also done this using the Frobenius automorphism, right?

b) Let $\alpha$ be a root of $f$ (in some algebraic closure of $\Bbb{F}_p$). Show that $\alpha + s$ is also a root for all $s \in \Bbb{F}_p$.

Let $\alpha^p - \alpha -1 =0$. Let $E$ be an algebraic closure of $\Bbb{F}_p$. Since $E$ has characteristic $p$, $(\alpha + s)^p = \alpha^p + s^p$. So we have,
$$(\alpha + s)^p - (\alpha + s) -1 = \alpha^p + s^p - \alpha -s -1 = s^p - s = 0.$$

c) Conclude that $f$ is irreducible over $\Bbb{F}_p$, for every $p$.

By b) and the fact that $\Bbb{F}_p$ has $p$ distinct elements, we know that the roots of $f$ are $\alpha, \alpha+1, ... , \alpha + p-1$. So if $K$ is a splitting field, we have $$x^p - x -1 = (x-\alpha)(x-(\alpha+1))...(x-(\alpha+p-1)).$$
Now let's assume that $f$ is reducible over $\Bbb{F}_p$. Then $f=gh$ for some $g$ and $h$ with degrees less than that of $f$. So $g$ and $h$ must be of the form $(x-(\alpha+s_1))...(x-(\alpha+s_k))$ where k is less than n. Let's say that g has degree 2, because the other cases are similar.
So $g =(x-(\alpha + s_i))(x-(\alpha + s_j))$ and the constant term for $g$ is,
$$(\alpha + s_i)(\alpha + s_j) = \alpha^2 + s_is_j\alpha + s_is_j.$$
Since $\Bbb{F}_p$ is a field, if $\alpha s_is_j$ is an element of $\Bbb{F}_p$ , then so is $((\frac{1}{s_is_j})(\alpha s_is_j) = \alpha$, a contradiction.
We can show by induction that if we multiply $(x-(\alpha+s_1))...(x-(\alpha+s_k)$, we get a term that looks like $s_1s_2...s_k\alpha$. So this is also true for $k&gt;2$.
Do you think that my answer is correct?
Thank you in advance
I like your idea very much, but I didn't understand how you managed to ignore the $\alpha^2$ term from your constant term. 

I have written up an answer to this question with a similar idea, where I look at the next to highest degree term of $g(x)$ (=the term of degree $k-1$). In your degree two example, this would be the linear term. Its coefficient is
$2\alpha+(s_i+s_j).$ As $s_i,s_j$ are in the prime field, and $2$ is invertible, we can, as in your argument, conclude that $\alpha\in\mathbb{F}_p$, which is a contradiction. 

By studying the lowest degree term, you get a lot of clutter from powers of $\alpha$. The degree $k-1$ term is IMHO easier to manage.
Another approach: 

Suppose $x^p-x -1 = gh$ for $g, h \in \mathbb F_p[x]$ with degrees $\geq 1$. Then taking a (formal) derivative, we have $$-1 = g'h + gh'$$ since $px^{p-1} = 0$ in $\mathbb F_p[x]$. Note that $g'$ and $h'$ cannot both be $0$, otherwise, $-1 = 0$. This implies the degree of the right hand side is at least $1$ (verify!). But the degree of $-1$ is $0$. Contradiction.