How $$\sum_{m=r}^{k-1}m(m-1)\ldots(m-r+1)=\sum_{m=r}^{k-1}\frac{m!}{(m-r)!};\quad r=1,2,\ldots$$ ?
i have $$\sum_{m=r}^{k-1}m(m-1)\ldots(m-r+1)=\sum_{m=r}^{k-1}\frac{m(m-1)\ldots[m-(r-1)][m-(r-2)]\ldots(3)(2)(1)}{[m-(r-2)]\ldots(3)(2)(1)}=\sum_{m=r}^{k-1}\frac{m!}{[m-(r-2)]!}$$
Where is the wrong ?
You wrote $$\sum_{m=r}^{k-1}m(m-1)\ldots(m-r+1)=\sum_{m=r}^{k-1}\frac{m(m-1)\ldots[m-(r-1)][m-(r-2)]\ldots(3)(2)(1)}{[m-(r-2)]\ldots(3)(2)(1)}.$$
The next term after $m-(r-1)$ should have been $m-r$. We are subtracting larger and larger numbers from $m$, until we get to $m-(m-1)=1$.