If $L$ contains a $n$-th root of $a\in K$, why does $K$ already contain a $n$-th root? I'm trying to solve the following problem from Algebra by Siegfried Bosch (english version below):

Es seien $m,n$ teilerfremde positive ganze Zahlen. Ist dann $L/K$ eine Körpererweiterung vom Grad $m$, so hat jedes Element $a\in K$, welches eine $n$-te Wurzel in $L$ besitzt, bereits eine $n$-te Wurzel in $K$.

My attempt at a translation:

Let $m,n$ be positive integers with $\mathrm{gcd}(m,n) = 1$. If $L/K$ is a field extension of degree $m$, then show that for any element $a\in K$ having an $n$-th root in $L$, there already is an $n$-th root of $a$ in $K$.

Since it's in the chapter, where trace and norm are defined, I'm guessing the norm will have to be used. So the question is: how?
The only (obvious) thing I can see:
If $b \in L$ is an $n$-th root of $a \in K$, then we have that 
$a^m = N_{L/K}(a) = N_{L/K}(b^n) = (N_{L/K}(b))^n$
and $N_{L/K}(b)\in K$. But I don't know whether (or how) this helps.
I don't seem to be able to come up with any good ideas, so I would appreciate some pointers. But please don't write down a full solution, if possible. I really would like to learn something here, so if you could give me just a hint, that would be great!
Thanks a lot.

Hint: You have shown that $a^m$ has an $n$th root in $K$. Also $a^n$ has an $n$th root in $K$.
Is the set of integers $\ell$ such that $a^\ell$ has an $n$th root in $K$ closed under ...?
Spoiler solution(for readers other than Sam, who solved his problem before this was added):

 $G=\{a^\ell\mid \ell\in\mathbf{Z}\}\cap {K^\times}^n$ is a multiplicative group. As a subgroup of a cyclic group it has to be cyclic itself, i.e. generated by some $a^t,t>0$. We saw that $a^m,a^n\in G$, so $t\mid m$, $t\mid n$ and the only possibility is $t=1$.

