On Selmer's polynomial Does this hold?

Let $p$ be an odd prime. If $\alpha$ is a root of $x^{p}-x-1$, prove that the ring of integers of $\mathbb{Q}[\alpha]$ is precisely $\mathbb{Z}[\alpha]$ and that this is a PID.

Is this true in general (without assuming $p$ to be prime, that is)? 
We know that $x^{n}-x-1$ is irreducible over $\mathbb{Q}$ for all $n$.
I checked this for $p=3$ and $p=5$, but in general I got pretty stuck... we have a formula for the discriminant of $x^{n}+ax+b$ but I wasn't able to finish. Any help?
Thanks in advance! 
 A: For $p=257$, the ring $\mathbb{Z}[\alpha]/(\alpha^p-\alpha-1)$ is not integrally closed. Let $K$ be the field of fractions of this ring.
Make the change of variable $\alpha = \beta+55$. Then the minimal polynomial of $\beta$ looks like $\beta^{257} + \cdots + a \beta^2+b \beta + c$ where $c$ is divisible by $59$ twice, $b$ is divisible by $59$ once and $a$ is not divisible by $59$. So the $59$-adic Newton polygon for this polynomial ends in a segment of length $2$ and slope $1$. More precisely, $a \equiv 27 \bmod 59$, $b \equiv 13 \cdot 59 \bmod 59^2$ and $c \equiv -1 \cdot 59^2 \bmod 59^3$ and $27 x^2 +13 x -1$ factors as $27(x-36)(x-40) \bmod 59$. So there is a place $\mathfrak{p}$ of $K$ lying above $59$ where $\alpha \equiv 55+36 \cdot 59 \mod \mathfrak{p}^2$ and another place $\mathfrak{q}$ where $\alpha \equiv 55+40 \cdot 59 \bmod \mathfrak{q}^2$. 
At this point, we know that the ring is not integrally closed: $59^2$ divides the discriminant of $x^{257}-x-1$, but there are no ramified places of $K$ above $59$. (The other factors of $x^{257}-x-1$ modulo $59$ are all squarefree, since the discriminant is only divisible by $59$ twice.) Actually constructing one of the missing integral elements would still be a bit of a pain.
