How $$\frac{1}{k}\sum_{m=r}^{k-1}\frac{m!}{(m-r)!}=\frac{(k-1)(k-2)\ldots(k-r)}{r+1};\quad r=1,2,\ldots$$
I have thought in two ways:
$$\frac{1}{k}\sum_{m=r}^{k-1}\frac{m!}{(m-r)!}=\frac{1}{k}\sum_{m=r}^{k-1}\frac{m(m-1)\ldots(m-r-1)(m-r)!}{(m-r)!}$$
Another $$\frac{1}{k}\sum_{m=r}^{k-1}\frac{m!}{(m-r)!}=\frac{1}{k}[\frac{r! }{0!}+\frac{(r+1)!}{1!}+\ldots+\frac{(k-1)!}{(k-1-r)!}]$$
Note that your result is equivalent to showing that
$$\frac{1}{k} \sum\limits_{m = r}^{k - 1} \frac{m!}{(m - r)!} = \frac{(k - 1)!}{(k - r - 1)! (r + 1)}$$
Fix an $r$ and induct on $k$; the base case, when $k = r + 1$ is immediate from:
$$\frac{1}{k} \sum\limits_{m = r}^r \frac{m!}{(m - r)!} = \frac{1}{r+1} \frac{r!}{0!} = \frac{r!}{r+1}$$
agreeing with the right hand side.
Now assume the result is true for some $k$; then
\begin{align} \frac{1}{k+1} \sum\limits_{m = r}^{(k + 1)-1} \frac{m!}{(m - r)!} &= \frac{1}{k + 1} \left(\sum\limits_{m = r}^{k - 1} \frac{m!}{(m - r)!} + \frac{k!}{(k - r)!}\right) \\ &= \frac{1}{k + 1} \left(k \cdot \frac{(k - 1)!}{(k - r - 1)! (r + 1)} + \frac{k!}{(k - r)!}\right) \\ &= \frac{1}{k + 1} \left(\frac{k! (k - r)}{(k - r)! (r + 1)} + \frac{k!(r + 1)}{(k - r)!(r + 1)}\right) \\ &= \frac{k!}{(k + 1)(k - r)!(r + 1)} \left( k - r + r + 1\right) \\ &= \frac{k!}{(k - r)!(r + 1)} \end{align}
as desired.