If $n^m\mid m^n$ and $k^n\mid n^k$, prove $k^m\mid m^k$, $m,n,k\in \mathbb{Z}^+$

Question

If $n^m\mid m^n$ and $k^n\mid n^k$, then $k^m\mid m^k$, $m,n,k\in \mathbb{Z}^+$

Aside from the definition of divisibility, can someone suggest theorems/facts that might be useful in proving this theorem?

Answer 1

Hint $\,\ \left(\dfrac{m^k}{k^m}\right)^n\!\! =\, \left(\dfrac{m^n}{\color{#c00}{n^m}}\right)^{\color{#c00}k} \left(\dfrac{\color{#c00}{n^k}}{k^n}\right)^{\color{#c00}m}\! \in\Bbb Z\ \color{#0a0}\Rightarrow\ \dfrac{m^k}{k^m}\in\Bbb Z\ $ by the Rational Root Test,

namely: $\quad x^n =\, j\in \Bbb Z\,$ has root $\ x\in\Bbb Q\ \color{#0a0}\Rightarrow\ x\in\Bbb Z,\ $ by the monic case of RRT (or, equivalently, an algebraic integer is rational iff it is an integer, i.e. $\,\Bbb Z\,$ is integrally closed; the proof in the linked answer works for any gcd domain, i.e. any integral domain where $\,\gcd(a,b)\,$ always exists).

Answer 2

$$n\ln m=\ln a+m\ln n,\ k\ln n=\ln b+n\ln k\\ \implies (n\ln m-\ln a)/m=(\ln b+n\ln k)/k\\ \implies kn\ln m-k\ln a=m\ln b+mn\ln k\implies n=(m\ln b+k\ln a)/(k\ln m-m\ln k)\\ \implies n= \ln_{\frac{m^k}{k^m}} \left(b^m a^k\right)$$ This gives $\left(\dfrac{m^k}{k^m}\right)^n={b^m}{a^k}$ which implies that $k^m|m^k$