= Solution
Put $k=\lfloor y^{1/n}\rfloor$. Because $y>0$, $k$ is a nonnegative <integer>, and
$$
k\leq y^{1/n}<k+1.
$$
The <function> $t\mapsto t^n$ is <strictly increasing> for $t\geq0$, so $k^n\leq y<(k+1)^n$. Both powers are <integers>. Taking the <floor function> therefore gives
$$
k^n\leq\lfloor y\rfloor<(k+1)^n.
$$
Taking nonnegative $n$th roots now gives $k\leq\lfloor y\rfloor^{1/n}<k+1$. Consequently
$$
\boxed{\left\lfloor\lfloor y\rfloor^{1/n}\right\rfloor=k=\left\lfloor y^{1/n}\right\rfloor.}
$$
The argument also covers $0<y<1$, when both outer <floors> are zero.
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