For an irreducible polynomial $p$ and root $\alpha$, $[F[\alpha]:F]$ = degree of $p$ I'm studying for an exam, and I couldn't find the proof for the following theorem in my notes:

For $p(x) \in F[x]$ irreducible and $\alpha$ a root of $p$ in some extension field, $[F[\alpha]:F] =$ degree of $p$

We haven't gotten far in field theory, and this is an undergraduate algebra course.
I'd appreciate it if anyone could provide a proof, a reference, or guidance for this.
 A: Hint. Note that $F[a]$ is a linear space over $F$ of dimension $n$, and basis $\{1,\alpha,\ldots,\alpha^{n-1}\}$, where $n$ the degree of the polynomial $p$, which is the one of minimum degree with root $\alpha$.  
A: If $d=\deg(p)$, the numbers $1,\alpha,\dots,\alpha^{d-1}$ are independent over $F$, as any vanishing linear combination gives us a polynomial of degree less than $d$ where $\alpha$ vanishes. Either this polynomial has all its coefficients $0$, or else $p$ and it have $\alpha$ as a common root and therefore their gcd is nontrivial, contradicting the irreducibility of $f$.
We can then assume that $1,\dots,\alpha^{d-1}$ are independent. This shows that, as a vector space, $F[\alpha]$ has dimension at least $d$ over $F$. 
On the other hand, any element of $F[\alpha]$ is a polynomial in $\alpha$ with coefficients from $F$. Given any such polynomial $q$, we can divide $q$ by $p$, say $q=pa+b$, where $a,b$ are polynomials in $F[x]$, and the degree of $b$ is strictly smaller than $d$. But then $q(\alpha)=b(\alpha)$, which shows that $q(\alpha)$ is a linear combination of $1,\dots,\alpha^{d-1}$.
This shows that $1,\dots,\alpha^{d-1}$ span $F[\alpha]$. Combining both facts, we see that $[F[\alpha]:F]=d$.
A: Suppose that $\deg(p) = n$.  You have to convince yourself that $F[\alpha]$ is $n$-dimensional, as an $F$-vector space.  Can you find a basis?  Consider the set
$$
\left\{ 1, \alpha, \alpha^2, \ldots, \alpha^{n - 1} \right\}.
$$
Why is the set linearly independent?  Why does it span $F(\alpha)$? 
