Theorem Let $p(x)$ be an irreducible polynomial in $F[x]$ and let $u$ be a root of $p(x)$ in an extension $E$ of $F$. Then if the degree of $p(x)$ is $n$, the set $(1,u,\ldots,u^{n-1})$ forms a basis of $F(u)$ over $F$.
What I did:
Note that $F(u)$ is by definition the smallest subfield of $E$ generated by $F$ and $u$. I proved that $F(u) = F[u]$ and I tried to prove the theorem, I found weird because since $F[u]$ is by definition the set $b_0+b_1u+\cdots+b_mu^m\in E$ such that $b_0+b_1x+\cdots+b_mx^m \in F[x]$ the basis should be $(1,u,\ldots,u^{n-1},u^n)$ with $n$ elements. where am I wrong? I need help.
Thanks