Exponent problem: if $m^n=n^{m^{n/m}}$ and $m,n>1$, prove that $m=n$. 
If $m^n=n^{m^{n/m}}$ and $m$ and $n$ are real numbers such that $m,n>1$, prove that $m=n$. 


I was playing around with exponents when I found this hypothesis. It appears to be true by plugging it into Desmos, but personally I can't find a mathematically rigorous proof for it. Any help is appreciated!
 A: 1. Case, $1< n<m$. Let $n>1$ be fixed and define $$f(m):=m^n-n^{m^{\frac nm}}.$$
Then, by some straight-forward but long calculations (this time verified with WolframAlpha ), we get $$f'(m)=nm^{n-1}+m^{\frac{n}{m}-2}\cdot n^{m^{\frac{n}{m}}+1}\cdot(\ln{m}-1)\cdot\ln(n).$$
Lemma. For $1\le n<m$, $f'(m)>0$.
Proof. If $m\geq e$, then the first summand is strictly positive and the second one is non-negative so we are done. Suppose thus that $n<m<e$. Then we note using $-1<\ln m - 1<0$ and $0\le\ln n<1$ that
$$f'(m)>nm^{n-1}-m^{\frac{n}{m}-2}\cdot n^{m^{\frac nm}+1}.$$
So $f'(m)>0$ is proven by:
\begin{align}
&\underbrace{m^{\frac nm-2} \cdot n^{m^{\frac nm}+1}}_{\le m^{-1}\cdot n^{m+1}}\le nm^{n-1} \\
\impliedby &n^{m+1} \le n\cdot m^n \\
\iff &n^m \le m^n \\
\iff &\frac{\ln(n)}n \le \frac{\ln(m)}m \\
\impliedby &n<m<e.\square
\end{align}
Since $f(n)=0$ and $f'(m)>0$ for all $n<m$, we conclude that $f(n)<0$ whenever $n<m$ so that there are no solutions to your equation for $n<m$.
2. Case, $1<m<n$. Now, let $m>1$ be fixed.
By taking $\log_m$ on both sides we see that your equation is equivalent to
\begin{equation}\tag 1\label 1 n=m^\frac{n}{m}\cdot\log_m(n).\end{equation}
Redefine $f(n):=m^\frac{n}{m}\cdot\log_m(n)-n$.
We have $f(1)=-1$ and $f(m)=0$.  
Note that \begin{equation}\tag 2\label 2 f'(n)=\frac{\partial}{\partial n} \left(m^\frac{n}{m}\cdot\log_m(n)-n\right)=m^{\frac nm}\cdot\left(\frac{\ln(n)}{m}+\frac{1}{n\ln(m)}\right)-1.\end{equation}
If we can prove that the expression in \eqref{2} is $>0$ for all $n>m,\infty$, we have proven that $f(n)>0$ for all $n>m$ so that there are no solutions when $n>m$.
For all $n>m$, we have \begin{equation}\tag 3\label 3m^{\frac nm}\cdot\left(\frac{\ln(n)}{m}+\frac{1}{n\ln(m)}\right)> \ln(n)+\frac 1n\frac{m}{\ln(m)}.\end{equation} We need to prove that the RHS of \eqref{3} is $>1$:
Both terms on the RHS of \eqref{3} are positive. If $n\geq e$ then we are done since $\ln(n)\geq 1$. 
Otherwise, note that $e>n>m$ so that $\frac{m}{\ln(m)}>e$ (the last function is decreasing and attains the value $e$ at $m=e$). It follows that $\frac1n\frac{m}{\ln(m)}>\frac1e\cdot e=1$ so we are done aswell.
It follows that the only solution is $n=m$.
