A conjecture formula: $\sum\limits_{n=1}^\infty \frac{\binom{mn}{n}}{n}\left(\frac{(m-1)^{m-1}}{m^m} \right)^n=m\log\left(\frac{m}{m-1}\right)$

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 ?

Let $z_m=(m-1)^{m-1}/m^m$. From this answer, we have \begin{align} F_m(z)&:=\sum_{n=0}^{\infty}\binom{mn}{n}\frac{z^n}{(m-1)n+1}=1+z\big(F_m(z)\big)^m, \\ G_m(z)&:=\sum_{n=0}^{\infty}\binom{mn}{n}z^n=\frac{F_m(z)}{m-(m-1)F_m(z)}. \end{align} Now $F_m(0)=1$ and $\color{blue}{F_m(z_m)=m/(m-1)}$ (yeah!), thus $$\sum_{n=1}^{\infty}\binom{mn}{n}\frac{(z_m)^n}{n}=\int_{0}^{z_m}\frac{G_m(z)-1}{z}\,dz,$$ and the substitution $w=F_m(z)$ (i.e. $z=(w-1)/w^m$) collapses it to $$\sum_{n=1}^{\infty}\binom{mn}{n}\frac{(z_m)^n}{n}=\color{blue}{m\int_1^{m/(m-1)}\frac{dw}{w}}=m\ln\frac{m}{m-1}.$$ [As a by-product, we get $\displaystyle\sum_{n=1}^{\infty}\binom{mn}{n}\frac{z^n}{n}=m\ln F_m(z)$.]