State/Rational roots of integers are integers

$$z = (-1)^{e_{-1} } \prod_{p \in P} p^{e_p},$$ $$q = (-1)^{f_{-1} } \prod_{p \in P} p^{f_p}.$$ Since $q^n = z$, we find that: $$q^n = (-1)^{n f_{-1} } \prod_{p \in P} p^{n \cdot f_p} = (-1)^{e_{-1} } \prod_{p \in P} p^{e_p} = z.$$ By the uniqueness of the canonical decomposition, the exponents of primes must match, and so: $$\forall p \in P, n \cdot f_p = e_p.$$ We claim that $f_p \geq 0$.

Hence $f_p \geq 0$, i.e., $f_p \in \NN$. It follows that the canonical decomposition of $q$ expressed $q$ as a product of integers.