How can $\sum_{k=i}^{n} \binom{k}{i}$ be written as
$$ \sum_{k=i}^{n} k(k-1)(k-2)\cdots(k-i+1) = \frac{(n+1)\cdot n\cdot (n-1)\cdots(n-i+1)}{i+1}$$
I cannot figure this one out, I have tried first proving by induction that the first formula is equal $\sum_{k=i}^{n} \binom{k}{i} = \binom {n+1}{i+1}$. But that did not seem to work for me, because I am confused by the fact that the summation index is equal to the lower limit, $k=i$. How could one approach this problem? And what does this equality mean?
Does the equality $$\sum_{k=i}^{n} \binom{k}{i} = \binom {n+1}{i+1}$$ hold because ${n+1\choose i+1}={k\choose i}+{k+1\choose i+1} ? $
As you said, if you could prove:
$$\sum_{k=i}^{n} \binom{k}{i} = \binom {n+1}{i+1}$$
Then we have:
$$\begin{align} \sum_{k=i}^{n} \frac{k!}{i!(k-i)!} &= \frac{(n+1)!}{(i+1)!(n-i)!}\\ \\ \sum_{k=i}^{n} \frac{k(k-1)...(k-i+1)}{i!} &= \frac{(n+1)n...(n-i+1)}{(i+1)!}\\ \\ \frac{1}{i!}\sum_{k=i}^{n} k(k-1)...(k-i+1) &= \frac{(n+1)n...(n-i+1)}{i!\cdot(i+1)}\\ \\ \sum_{k=i}^{n} k(k-1)...(k-i+1) &= \frac{(n+1)n...(n-i+1)}{i+1} \end{align}$$