Equality of Integers to the Power of Each Other/Proof 1

Proof
We have:

, let $m > n$.

We have that $m, n \in \N$.

Hence:
 * $\log_n m \in \N \implies m = n^k$

where $k \in \N_{>1}$.

Hence:
 * $n^k = k n$

For $n \ne 0$ the solution reads


 * $n = k^\frac 1 {k - 1}$

Define $t = k - 1$.

We have:

To check for intermediate maximum consider the first derivative:


 * $\dfrac {\d n} {\d k} = \dfrac {k^{\frac 1 {k - 1} } } {k - 1} \paren {\dfrac 1 k - \dfrac {\ln k} {k - 1} }$

Our desired solution constrains the prefactor to be positive.

The term in brackets vanishes only for $k = 1$.

Hence for $k > 1$:
 * there is no extremum
 * $\map n k$ is monotonically decreasing
 * $1 < n < e$.

Hence the only natural number solution is $n = 2$.

The only $k$ that satisfies this is $k = 2$.

Therefore:
 * $n = 2$
 * $m = 2^2 = 4$