Prove $\alpha = [X] \in L = \mathbb F_p[X]/\langle \pi \rangle$ is a primitive n'th root of unity in $L$ and this implies $p^d \equiv 1 \mod n$ [My rephrasing, JL]
Let $p$ be a prime and let $n$ a positive integer coprime to $p$. Let further $\Phi_n(X)$ denote the $n$th cyclotomic polynomial, and let $\pi\in\Bbb{F}_p[X]$ be an irreducible factor of degree $d$ of $\Phi_n(X)$. Let $L=\Bbb{F}_p[X]/\langle \pi\rangle$ be the quotient field. Consider the element $\alpha=[X]\in L$. Show that it is a primitive $n$th root of unity in $L$. Further explain, why this implies that $p^d\equiv1\pmod n$.

Prove $\alpha = [X] \in L = \mathbb F_p[X]/\langle \pi\rangle$ is a primitive n'th root of unity in $L$ and this implies $p^d \equiv 1 \mod n$, where $\pi \in \mathbb F_p[X]$ is irreducible and $\pi \mid \phi_n(X) \in \mathbb F_p[X]$.
I know $p \nmid n$ and $\phi_n(X)$ denote the n'th cyclotomic polynomial.
I've proven that $L$ is a field with $p^d$ elements. 
$\pi \mid \phi_n(X)$ allows us to write $\pi g=\phi_n(X)$ for some $g \in \mathbb F_p[X]$. $\langle a_d^{-1} \pi \rangle = \langle \pi \rangle$ tells us that $\alpha^d= 1\in L$. Also $[\phi_n(X)] = [\pi g]$, but I can't proceed from here.
 A: Extended hints (that's my style):
We are first to prove that $\alpha^n=1_L$, and that $n$ is the smallest positive integer with this property.


*

*At the level of cosets of polynomials the equation $\alpha^n=1$ means that the cosets $X^n+\langle \pi\rangle$ and $1+\langle\pi\rangle$ are the same. In other words, the coset of their difference, $X^n-1$ should be the zero coset. Yet in other words, we first need to show that the polynomial $X^n-1$ is divisible by $\pi$. Why is that the case?

*Next we need to show that if $0<\ell<n$, then $\alpha^\ell\neq1_L$. As in the preceding bullet point, this is equivalent to showing that $X^\ell-1$ is not divisible by $\pi$ for any such $\ell$. Note that we only need to worry about the cases where $\ell\mid n$. This is because the first bullet implies that the order of $\alpha$ is a factor of $n$. This step is actually a tad tricky, because you need the fact that $\gcd(n,p)=1$. That condition implies that $X^n-1$
has no multiple factors (how?), and you will need that here (why?)


The other part of the exercise is the claim that $n\mid p^d-1$. This is now easy if you remember the following bits:


*

*The group $L^*$ is cyclic of order $p^d-1$, because the multiplicative grops of finite fields are always cyclic.

*Lagrange's theorem from the theory of finie (cyclic) groups. Actually we already used this in the second bullet while showing that we only need to look at the cases $\ell\mid n$.

