Proof existence of field extension of $\mathbb{F}_p$ containing the $r$-th primitive root of unity I have to show the following:

Let $p$ be a prime and $r \in \mathbb{N}$ with $\gcd(r,p)=1$. Prove the existence of a field extension $E$ of $\mathbb{F}_p$ which contains an $r$-th primitive root of unity.

I don't know how to start here.
Any help is appreciated.
Greetings
 A: Consider the residues modulo $r$ of $p, p^2, p^3, \dots, p^{r-1}$. 
Claim: $p^m \equiv 1 \bmod r$ for some $m, 1 \leq m \leq r-1$.
The proof is by contradiction.  All
$r-1$ residues belong to the set $\{1,2,\ldots, r-1\}$, and so if
none of them equals $1$, then two residues must have the same
value and thus $r$ is a divisor of $p^i - p^j = p^j(p^{i-j}-1)$.
Since $\gcd(r, p^j) = 1$, $r$ must be a divisor of $(p^{i-j}-1)$.
that is, $p^{i-j} \equiv 1 \bmod r$ in contradiction of the 
assumption that none of the residues equals  $1$.  $\quad\qquad\Box$
Let $m$ denote the smallest positive integer such that $r$ 
divides $p^m-1$.  There exist irreducible polynomials of 
degree $m$ in $\mathbb F_p[x]$ (for the exact number,
see, for example, here), 
and if $g(x)$ is such a polynomial, then 
$E = \mathbb F_p[x]/g(x)$ is an extension field 
of $\mathbb F_p$. $|E| =p^m$ and its nonzero elements
constitute a cyclic group of order $p^m-1$ under multiplication.
This group has elements of all orders $n$ that divide $p^m-1$.
Since $r$ is a divisor of $p^m-1$, $E$ contains a primitive  $r$-th root
of unity, which is what you have to show. $\quad\qquad\Box$
A: Note that $X^r-1$ is separable over $\mathbb F_p$, and let $E$ be a splitting field for $X^r-1$ over $\mathbb F_p$.
EDIT A. If you don't know the notion of splitting field, you can proceed as follows. Let $E/\mathbb F_p$ be an extension containing $m$ roots of $X^r-1$, with $m$ maximum. Assume by contradiction $m < r$. Then $X^r-1$ has a factor $f$ of degree at least $2$ which is irreducible over $E$. Form the extension $E[X]/(f)$ to get a contradiction. 
Of course, the key point is that the roots of $X^r-1$ in any extension of $\mathbb F_p$ are simple. (Do you see why?)
EDIT B. The following fact has been implicitly used: a finite subgroup of the multiplicative group of a field is cyclic. So, once you have $r$ distinct roots of $X^r-1$ in $E$, you know that one of them (at least) is primitive.
EDIT C. The argument in Edit A shows this: Let $K$ be a field and $f\in K[X]$ a polynomial of degree $n$. Then there is an extension $L/K$ which contains $n$ roots of $f$, counted with their multiplicities. It's a very basic fact.
