A conjecture formula: $\sum\limits_{n=1}^\infty \frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n=m\log\left(\frac{m}{m-1}\right)$ By the help of Mathematica numeral calculations, I find the following formula holds
$$\sum\limits_{n=1}^\infty \frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n=m\log\left(\frac{m}{m-1}\right)\quad ?$$
$m>1$ is a positive integer. But I can't prove it. 

This is hardly a starter,  but the following representation showing  some commonalities of  LHS and  RHS might be useful.
The RHS  can be  written as
\begin{align*}
m\log\left(\frac{m}{m-1}\right)&=m\log\left(\frac{1}{1-\frac{1}{m}}\right)\\
&=-m\log\left(1-\frac{1}{m}\right)\\
&\,\,\color{blue}{=m\sum_{n=1}^\infty \frac{1}{nm^n}}
\end{align*}
The  LHS  can  be written as
\begin{align*}
\sum_{n=1}^\infty&\frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n\\
&=\sum_{n=1}^\infty\frac{1}{nm^n}\binom{mn}{n}\left(\frac{(m-1)^{m-1}}{m^{m-1}}\right)^n\\
&=\sum_{n=1}^\infty\frac{1}{nm^n}\binom{mn}{n}\left(1-\frac{1}{m}\right)^{n(m-1)}\\
&\,\,\color{blue}{=m\sum_{n=1}^\infty\frac{1}{nm^n}\binom{mn-1}{n-1}\left(1-\frac{1}{m}\right)^{n(m-1)}}
\end{align*}
