Proving that $\frac {(n+1)n(n-1)...(n-i)}{i+1}$ can be used as the summ of $k^h$ I have previously proved that $$\sum_{k=i}^{n} k(k-1)(k-2)\cdots(k-i+1) = \frac{(n+1)\cdot n\cdot (n-1)\cdots(n-i+1)}{i+1}$$
From here I am supposed to show that the formula above can be used to compute the sum of $k^h$, for example, $\sum_{k=1}^{n}k^2$ or $\sum_{k=1}^{n}k^3$, starting with $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$$
I cannot figure out the connection between these and do not know where to start...

Yes, it can be used.
Example with $\sum_{k=1}^{n}k^2$
$\sum_{k=1}^{n}k^2$
$=\sum_{k=1}^{n}k(k-1+1)$
$=(\sum_{k=1}^{n}k(k-1)) + \sum_{k=1}^{n}k$
$=(\sum_{k=2}^{n}k(k-1)) + \sum_{k=1}^{n}k$ ( means that $i = 2$)
$=\frac{(n+1)n(n-1)}{3} + \frac{n(n+1)}{2}$
$=\frac{(n+1)n(2(n-1) + 3)}{6}$
$=\frac{(n+1)n(2n+1)}{6}$

Example with $\sum_{k=1}^{n}k^3$
$\sum_{k=1}^{n}k^3$
$=\sum_{k=1}^{n}k(k-1+1)(k-2+2)$
$=\sum_{k=1}^{n}k((k-1)(k-2) + 2(k-1)+(k-2)+2)$
$=\sum_{k=1}^{n}k((k-1)(k-2) + 3k-2)$
$=\sum_{k=1}^{n}k(k-1)(k-2) + 3\sum_{k=1}^{n}k^2 - 2\sum_{k=1}^{n}k$
$=\sum_{k=3}^{n}k(k-1)(k-2) + 3\sum_{k=1}^{n}k^2 - 2\sum_{k=1}^{n}k$ ( means that $i=3$)
$=\frac{(n+1)n(n-1)(n-2)}{4} + 3\frac{n(n+1)(2n+1)}{6} - 2\frac{n(n+1)}{2}$
$=\frac{n^2 (n + 1)^2}{4}$

Same way in higher power
