= Solution
Now let $n=m=p$ be <prime> and put $\alpha=\sqrt[p]{p}$ and $A=\mathbb Z[\alpha]$. The <Eisenstein criterion> applied at $p$ gives $[\mathbb Q(\alpha):\mathbb Q]=p$, while the <discriminant of elements of a number field> of its power basis is
$$
|\operatorname{disc}(1,\alpha,\ldots,\alpha^{p-1})|
=|N(f'(\alpha))|
=|N(p\alpha^{p-1})|
=p^{2p-1},
$$
where $f(X)=X^p-p$. By the <discriminant-index formula for an integral lattice>, every prime dividing $[\mathcal O_L:A]$ must therefore be $p$.
Suppose $A\ne\mathcal O_L$. The finite <abelian group> $\mathcal O_L/A$ is then a nontrivial <finite p-group>, so it contains an element of order $p$. Consequently there is an $x\in\mathcal O_L\setminus A$ such that
$$
x=\frac{g(\alpha)}p,
\qquad
g(X)=\sum_{k=0}^{p-1}c_kX^k\in\mathbb Z[X],
$$
with some $c_k$ not divisible by $p$.
The polynomial $X^p-p$ is Eisenstein, so <total ramification from an Eisenstein polynomial> gives a unique prime ideal $P$ above $p$ with normalized <discrete valuation>
$$
v_P(\alpha)=1,
\qquad
v_P(p)=p.
$$
Let $k_0$ be the least index for which $p\nmid c_{k_0}$. The term $c_{k_0}\alpha^{k_0}$ has valuation $k_0<p$; every earlier term has valuation at least $p$, and every later term with coefficient prime to $p$ has a distinct, larger valuation. The non-Archimedean valuation therefore gives
$$
v_P(g(\alpha))=k_0<p,
\qquad
v_P(x)=k_0-p<0.
$$
This contradicts $x\in\mathcal O_L$, since an <algebraic integer> has nonnegative valuation at every <prime ideal>. Thus $A=\mathcal O_L$, and
$$
\boxed{\mathcal O_L=\mathbb Z[\sqrt[p]{p}]}.
$$
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