If $\Phi(x)=x^{p}$ is the Frobenius automorphism of $\mathbb{F}_{q}$, where $q = p^d$ , what is the smallest integer $\alpha$ such that $\Phi^{\alpha}=\text{Id}$?
Since the order of the group $\mathbb{F}_q^*$ is $q-1$, so we know that the elements of this group satisfy $x^{q-1} = 1 $, thus $$x^q = x$$ and therefore $$x^{p^d} = x, \text{ by definition of the Frobenius automorphism we know that's equivalent to saying that } \\ \phi^d = x. $$ Thus the order of $\phi$ divides $d$, but how do I show that the order is exactly $d$?
If there was $k<d$ such that $\phi^k = \mathrm{Id}$, then every $x\in \mathbb{F}_q$ would satisfy:
$$x^{p^k}-x = 0.$$
That is, they would all be roots of a polynomial with degree $p^k$ and coefficients in $\mathbb{F}_p$. Such a polynomial can only have at most $p^k$ distinct roots, and we know that $\mathbb{F}_q$ has $p^d>p^k$ elements. This results in contradiction, so $d$ is the order of $\phi$.