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...
 A: It is convenient to use the falling factorial notation
\begin{align*}
k^{\underline{i}}:=k(k-1)\cdots (k-i+1)
\end{align*}
With this notation OPs identity can be written as
\begin{align*}
\sum_{k=i}^nk^{\underline{i}}=\frac{(n+1)^{\underline{i+1}}}{i+1}
\end{align*}

and since summands with $1\leq k<i$ are zero we can also write
\begin{align*}
\color{blue}{S_i(n):=\sum_{k=1}^nk^{\underline{i}}=\frac{(n+1)^{\underline{i+1}}}{i+1}}\tag{1}
\end{align*}
We can now iteratively use $S_i(n), i=1,2,3,\ldots $ to calculate
\begin{align*}
\sum_{k=1}^nk^i
\end{align*}

Case $i=1$:
Since $k=k^{\underline{1}}$ we obtain from (1)
\begin{align*}
\color{blue}{\sum_{k=1}^n k}=S_1(n)=\frac{(n+1)^{\underline{2}}}{2}
\color{blue}{=\frac{1}{2}\left(n+1\right)n}\tag{2}
\end{align*}
Case $i=2$:
Since $k^2=k(k-1)+k=k^{\underline{2}}+k^{\underline{1}}$ we obtain from (1) and (2)
\begin{align*}
\color{blue}{\sum_{k=1}^n k^2}&=S_2(n)+S_ 1(n)=\frac{(n+1)^{\underline{3}}}{3}+\frac{(n+1)^{\underline{2}}}{2}\\
&=\frac{(n+1)n(n-1)}{3}+\frac{(n+1)n}{2}\\
&\,\,\color{blue}{=\frac{1}{6}\left(2n^2+3n^2+n\right)}
\end{align*}
This way we can iteratively calculate $\sum_{k=1}^n k^i, i=3,4,\ldots$. Here it is useful to know that
\begin{align*}
k^i=\sum_{j=0}^i\begin{Bmatrix} i \\ j \end{Bmatrix} k^{\underline{j}}
\end{align*}
with $\begin{Bmatrix} i \\ j \end{Bmatrix}$ the Stirling numbers of the second kind.
Note: The sum formula (1) is a discrete analogon to
\begin{align*}
\int_{0}^xt^i\,dt=\frac{x^{i+1}}{i+1}
\end{align*}
which is a main theme in this answer .
A: 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
