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. 

Too long for comments.
Using another CAS, I have not been able to obtain the rhs (except for $m=2$) but numerically the results do agree with your conjecture (checked up to $m=20$).
Considering
$$f_m=\sum\limits_{n=1}^\infty \frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n$$ running cases, what I obtained is
$$f_3=\frac{2 ^2}{3^2} \, _4F_3\left(1,1,\frac{4}{3},\frac{5}{3};\frac{3}{2},2,2;1\right)$$
$$f_4=\frac{3^3}{4^3} \,
   _5F_4\left(1,1,\frac{5}{4},\frac{6}{4},\frac{7}{4};\frac{4}{3},\frac{5}{3},2,2;1
   \right)$$
$$f_5=\frac{4^4}{5^4} \,
   _6F_5\left(1,1,\frac{6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5};\frac{5}{4},\frac
   {6}{4},\frac{7}{4},2,2;1\right)$$
$$f_6=\frac{5^5}{6^5}\,
   _7F_6\left(1,1,\frac{7}{6},\frac{8}{6},\frac{9}{6},\frac{10}{6},\frac{11}{6};\frac
   {6}{5},\frac{7}{5},\frac{8}{5},\frac{9}{5},2,2;1\right)$$
$$f_7=\frac{6^6}{7^6}\,
   _8F_7\left(1,1,\frac{8}{7},\frac{9}{7},\frac{10}{7},\frac{11}{7},\frac{12}{7},
   \frac{13}{7};\frac{7}{6},\frac{8}{6},\frac{9}{6},\frac{10}{6},\frac{11}{6},2,2;1
\right)$$ which, as written, reveal very clear patterns.
$$\color{blue}{f_m=\frac{(m-1)^{m-1}}{m^{m-1}}\,
   _{m+1}F_m\left(1,1,\frac{m+1}m,\cdots,\frac{2m-1}m;\frac m{m-1},\cdots,\frac {2m-3}{m-1},2,2;1\right)}$$ 
Trying on Wolfram Cloud, I obtained the same results but no simplification at all. Surprising, isn't it ?
