= Solution
Put $\alpha=\sqrt[n]{m}$. It is an <algebraic integer> because it satisfies the monic polynomial $X^n-m$. Choose a <prime number> $q$ dividing the <square-free integer> $m$, and choose a <prime ideal> $\mathfrak p$ of $\mathcal O_L$ above $q$. From
$$
(\alpha)^n=(m)
$$
and <prime ideal factorization>, the $\mathfrak p$-adic exponents satisfy
$$
n v_{\mathfrak p}(\alpha)=v_{\mathfrak p}(m)=e_{\mathfrak p},
$$
because $q$ occurs to exponent one in $m$. Hence $n$ divides the <ramification index of a prime ideal> $e_{\mathfrak p}$. If $d=[L:\mathbb Q]$, then $e_{\mathfrak p}\leq d$, while the equation $\alpha^n=m$ gives $d\leq n$. Therefore
$$
n\leq e_{\mathfrak p}\leq d\leq n,
$$
so
$$
\boxed{[L:\mathbb Q]=n}.
$$
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