Question about cancellations in a series resulting in $\frac {m+1}{m-k+1}$ I have the following series:
$$\displaystyle\sum_{i=1}^{k+1}iP_i = 1\left(1-\frac k m\right)+2\frac k m \left(1-\frac{k-1}{m-1}\right)+3\frac k m \frac {k-1}{m-1}\left(1-\frac{k-2}{m-2}\right)\\+...+(k+1)\frac k m \frac {k-1}{m-1}...\left(\frac{k-(k-1)}{m-(k-1)}\right) =\frac {m+1}{m-k+1}$$
I don't get how was the cancellation done here to get to that expression? How were the increasing $i$s canceled?
Trying to see the pattern with numbers, plugging $k=3, m=7$:
$1-\frac 3 7 +2\frac 3 7(1-\frac 2 6)+3\frac 3 7 \frac 2 6 (1-\frac 1 5)+4\frac 3 7 \frac 2 6 \frac 1 5 = \frac 4 7 (1+ 1 +\frac 3 5+ \frac 1 5 )=1.6$ which is correct but I still don't see the pattern. 
 A: This looks harder than it is.
Notice that your series can be written as:
\begin{align}
1 - & \frac{k}{m} + 2\frac{k}{m} - 2\frac{k}{m} \cdot \frac{k-1}{m-1}+3\frac{k}{m}\cdot \frac{k-1}{m-1}-3\frac{k}{m}\cdot \frac{k-1}{m-1}\cdot \frac{k-2}{m-2}+\dots+ \\ & (k-1)\cdot \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot \frac{3}{m-k+3} - (k-1)\cdot \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot \frac{3}{m-k+3} \cdot \frac{2}{m-k+2} + \\ & \dots + k \cdot \frac{k}{m}\cdot \frac{k-1}{m-1}\cdot \dots \cdot \frac{2}{m-k+2} - k \cdot \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot \frac{2}{m-k+2} \cdot \frac{1}{m-k+1} + \dots + \\ & (k+1) \cdot \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot \frac{2}{m-k+2} \cdot \frac{1}{m-k+1} \text{(last element equals $0$ as you noticed)} = \\ & 1 + \frac{k}{m} + \frac{k}{m} \cdot \frac{k-1}{m-1} + \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \frac{k-2}{m-2} + \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot \frac{3}{m-k+3}+ \\ & \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot \frac{3}{m-k+3} \cdot \frac{2}{m-k+2} + \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot \frac{2}{m-k+2} \cdot \frac{1}{m-k+1} = \\ & 1+ (\frac{k}{m} \cdot (1+\frac{k-1}{m-1} \cdot (1+\frac{k-2}{m-2} \cdot ( \dots \cdot(1+\frac{2}{m-k+2} \cdot (1+\frac{1}{m-k+1}))\dots)))) = \\ &  1+ (\frac{k}{m} \cdot (1+\frac{k-1}{m-1} \cdot (1+\frac{k-2}{m-2} \cdot ( \dots \cdot(1+\frac{2}{m-k+2} \cdot (\frac{m-k+2}{m-k+1}))\dots)))) = \\ & 1+ (\frac{k}{m} \cdot (1+\frac{k-1}{m-1} \cdot (1+\frac{k-2}{m-2} \cdot ( \dots \cdot (\frac{3}{m-k+3}\cdot(\frac{m-k+3}{m-k+1} )\dots)))) = \dots = \\ & 1+ \frac{k}{m-k+1} = \frac{m+1}{m-k+1}
\end{align}
