Finite Field Extension Suppose $E/F$ is a field extension of degree $n$. Does it follow that $E = F(a_{1}, a_{2}, \ldots, a_{n})$ for some $a_{i} \in E$? I feel like this is true, but I'm getting confused with all the definitions.
 A: Yes, the statement that $E=F(a_1,\ldots,a_n)$ for some $a_i\in E$ is implied by the statement that $[E:F]=n$. Recall that $[E:F]=n$ is, by definition, the statement that $E$ is an $n$-dimensional vector space over $F$. Thus, there exists a basis for this vector space consisting of $n$ elements, that is, there exist $a_1,\ldots,a_n\in E$ such that any $b\in E$ has
$$b=c_1a_1+\cdots+c_na_n$$
where the $c_i\in F$. Thus, we will have $E=F(a_1,\ldots,a_n)$.
However, you should note that it might not be the case that all the $a_i$'s are necessary to generate $E$ as a field over $F$. For example, if $E=\mathbb{Q}(\sqrt[n]{2})$ and $F=\mathbb{Q}$, then $[E:F]=n$, and a basis for $E$ as a vector space over $F$ is 
$$1,\sqrt[n]{2},\ldots,(\sqrt[n]{2})^{n-1}$$
so that $E=F(1,\sqrt[n]{2},\ldots,(\sqrt[n]{2})^{n-1})$, but we actually already had $E=F(\sqrt[n]{2})$ - that is, $E$ can be generated over $F$ by a single element. 
A: E being a degree n extension of F means, that E as a vectorspace over F has dimension n. 
So there are n elements of E which generate E as a vectorspace over F. These elements will obviously also generate E as a field over F.
A: Yes.  Using the fact that $F(a,b) = F(a)(b)$, you can prove this by induction on the degree $[E:F]$.
Your bound on the number of generators in terms of the degree is sharp only in the trivial case $[E:F] = 1$, i.e., $E = F$.  Since every time you introduce another generator the degree goes down by a factor of $d$ for some divisor $d$ of $n$ with $d > 1$, you can see that in fact you can get $E$ by adjoining at most $\omega(n)$ elements to $F$, where $\omega(n)$ is the number of prime factors of $n$, counting multiplicity.  So, for instance, $\omega(n) \leq \log_2(n)$ for all $n > 1$.
Even this bound could only be sharp if $n = p^k$ and $F$ has characteristic $p$.  But by now I am alluding to questions that no one has asked, so I'll stop here...
(Added: Of course I agree that one can just pick a basis of $E/F$ and remark that finitely generated as an $F$-module implies finitely generated as an $F$-algebra implies finitely generated as a field extension of $F$.  I just decided to give a different argument, for whatever reason...)
