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. 
 A: 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 ?
A: 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 $\displaystyle\sum_{n=1}^{\infty}\binom{mn}{n}\frac{z^n}{n}=m\ln F_m(z)$.]
A: 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*}
A: Here's my attempt at closed form. Not an answer definitely, but might be of use.
First, we simplify (and generalize) the problem by defining a two variable series:
$$S(x,y)=\sum_{n=1}^\infty \binom{nx}{n} \frac{y^n}{n}$$
In the OP we have:
$$y=\frac{1}{x} \left(1-\frac{1}{x} \right)^{x-1}$$
Now we assume $x \in \mathbb{R}$, but $x \notin \mathbb{Z}$ (we can get back to the whole numbers by continuity arguments), then we can represent the binomial coefficient in the following way:
$$\binom{nx}{n}= \frac{x}{\pi}\sin[\pi n (1-x)] B[nx,n(1-x)] $$

Here's the main problem: all the (real) integral representations of the Beta function rely on both the arguments being positive. But it would only be the case if $0<x<1$, which doesn't fit the OP. However, I have checked the original series, and despite giving complex values, the closed form still seems to work for $|x|<1$, so I will consider this case first.

Attempt 1
We have:
$$B[nx,n(1-x)]=\int_0^1 t^{n x-1} (1-t)^{n(1-x)-1}dt=\int_0^1 \left[t^x (1-t)^{1-x} \right]^n \frac{dt}{t(1-t)}$$
$$\sin[\pi n (1-x)]=\frac{1}{2i} \left(e^{\pi i (1-x) n}-e^{-\pi i (1-x) n} \right)$$
Then we can write:
$$S(x,y)=\frac{x}{2 i \pi} \int_0^1  \frac{dt}{t(1-t)} \sum_{n=1}^\infty \left(e^{\pi i (1-x) n}-e^{-\pi i (1-x) n} \right) \left[t^x (1-t)^{1-x} \right]^n  \frac{y^n}{n} $$
We also need $|y|<1$, which doesn't seem to work for $|x|<1$ if we define $y$ as in the original series, however let's forget about that for now and sum the series formally:

$$S(x,y)=-\frac{x}{2 \pi i} \int_0^1  \frac{dt}{t(1-t)} \log \frac{1-e^{\pi i (1-x)} t^x (1-t)^{1-x} y }{1-e^{-\pi i (1-x)} t^x (1-t)^{1-x} y}, \qquad 0<x<1$$

If we set $y=\frac{1}{x} \left(1-\frac{1}{x} \right)^{x-1}$, then the closed form $-x \log \left( 1-\frac{1}{x} \right)$ works numerically, as in, the real and imaginary parts are the same. Though I don't know how to prove it for the integral either.
Attempt 2
For another try we could turn to Gamma functions, which are better defined:
$$\binom{nx}{n}= \frac{x}{\pi}\sin[\pi n (1-x)] \frac{\Gamma(nx) \Gamma(n(1-x))}{(n-1)!} $$

To work with the usual integral representation of the Gamma function, we again have to restrict ourselves to $0 <x <1$, however, as we will see, it will allow us to consider $|y|>1 $ as well.

$$\Gamma(nx) \Gamma(n(1-x))=\int_0^\infty \int_0^\infty u^{nx} v^{n(1-x)} e^{-u-v} \frac{du dv}{u v}$$
So we have:
$$S(x,y)=\frac{x}{2\pi i} \int_0^\infty \int_0^\infty e^{-u-v} \frac{du dv}{u v} \sum_{n=1}^\infty \left(e^{\pi i (1-x) n}-e^{-\pi i (1-x) n} \right) [u^x v^{1-x}]^n  \frac{y^n}{n!}$$
Summation gives us:
$$S(x,y)=\frac{x}{2\pi i} \int_0^\infty \int_0^\infty e^{-u-v} \frac{du dv}{u v} \left(\exp \left[y e^{\pi i (1-x)} u^x v^{1-x} \right]-\exp \left[y e^{-\pi i (1-x)} u^x v^{1-x} \right] \right) $$
Getting rid of complex numbers:
$$S(x,y)=\frac{x}{\pi} \int_0^\infty \int_0^\infty e^{-u-v} \exp \left[y \cos (\pi  (1-x)) u^x v^{1-x} \right] \sin \left[y \sin (\pi  (1-x)) u^x v^{1-x} \right] \frac{du dv}{u v}$$
This integral seems to work as well, though numerical evaluation is very difficult.
A: EDIT: This answer is incorrect, due to a mistaken bound on the binomial coefficient. In fact, $\binom{mn}{n}\leq 2^{mn}$ so of course it has exponential growth...

The formula cannot hold, since the left side is a divergent series! Indeed, for every fixed $m$ the $\binom{mn}{n}$ grows faster than exponentially in $n$, since it is greater than $((m-1)n)^n$. But this means it outgrows the reciprocal of the rest of the summand...
