Show that if every element of $K$ is a root of a polynomial in $F[x]$ of degree less than or equal to $n$, then $K$ is a simple extension of $F$ of degree less than or equal to $n$.
I know by the primitive element theorem that if $K$ is a finite extension of $F$ then $K$ is a simple extension so $K = F(a)$ for some $a \in K$ and since deg(min$_F(a)$) $\leq n$ we have $|F(a) : F| \leq n$. If $F$ is finite then its clear that $|K:F| < \infty$. My only problem is I can't figure out how to get $|K : F| < \infty$ if $F$ is infinite.
I was thinking about supposing for the sake of contradiction that $|K: F| = \infty$ and then defining $F = F_1 \subseteq F_2 \subseteq ... \subseteq F_n = K$ such that each $F_i$ is the intermediate field obtained by adjoining all the elements of $K$ that are the root of a polynomial of degree less then or equal o $i$ over $F$. Since $K: F| = \infty$ we must have $|F_{i + 1} : F_i| = \infty$ for at least one integer $i$. But I am out of ideas. Any help is appreciated.
If you assume that $K$ is separable and $[K:F]$ is infinite.
Let $x_1\in K-F$, $[F(x):F]\leq n$, there exists $x_2$ which is not in $F(x_1)$, consider $F(x_1,x_2)$ it is finite and $[F(x_1,x_2):F]>[F(x_1):F]$ recursively, we construct a finite extension $F_i=F(x_1,...,x_n)$ such that $[F(x_1,...,x_{n+1}):F]>[F(x_1,...,x_n):F]$, primitive element theorem $F_i=F(y_i)$ contradiction since $deg(y_i)<n$ for every $y_i$.