How do I prove the following identity:
$$\sum_{r=1}^{m} \frac{(m+1)(r-1)m^{r-1}}{r {m \choose r}} = m^m -1$$
I tried using the method shown in this answer to get the following integral:
$$\frac{m+1}{m} \int_{0}^{1}(\frac{m^{m+1}(1-t)^{-1}t^{m-1}}{t(m+1)-1} + \frac{m(1-t)^{m-1}}{(t(m+1)-1)^2} + \frac{m^{m+1}t^m(1-t)^{-1}}{(t(m+1)-1)^2}) dt$$
But I have no idea on how to proceed from here.
This sum can be found using elementary methods (and for once, "elementary" does actually mean "simple"). We have: $$\frac{m^r}{\binom mr} - \frac{m^{r-1}}{\binom m{r-1}} = \frac{m^r}{\binom mr} - \frac{m^{r-1}}{\frac{r}{m-r+1}\binom mr} = \frac{m^r}{\binom mr}\left(1 - \frac{m-r+1}{mr}\right) = \frac{(m+1)(r-1) m^{r-1}}{r \binom mr}$$ which is exactly the summand in your identity.
Therefore the sum telescopes: $$\sum_{r=1}^m \frac{(m+1)(r-1) m^{r-1}}{r \binom mr} = \sum_{r=1}^m \left(\frac{m^r}{\binom mr} - \frac{m^{r-1}}{\binom m{r-1}}\right) = \frac{m^m}{\binom mm} - \frac{m^0}{\binom m0} = m^m-1.$$
