Sums of rising factorial powers Doodling in wolfram, I found that
$$
\sum^{k}_{n=1}1=k
$$
The formula is pretty obvious, but then you get
$$
\sum^{k}_{n=1}n=\frac{k(k+1)}{2}
$$
That is a very well known formula, but then it gets interesting when you calculate
$$
\sum^{k}_{n=1}n(n+1)=\frac{k(k+1)(k+2)}{3}\\
\sum^{k}_{n=1}n(n+1)(n+2)=\frac{k(k+1)(k+2)(k+3)}{4}
$$
And so on. There is an obvious pattern that I really doubt is a coincidence, but I have no idea how to prove it in the general case. Any ideas?

Partial sums of sequences $a_n$ that can be expressed as polynomials in $n$ are easily found using discrete calculus.
We start with the discrete version of the Fundamental Theorem of Integral Calculus:
\begin{align*}
    \sum\limits_{n=1}^k a_n &= \sum\limits_{n=1}^k (\Delta b)_n \\
         &= (b_2 - b_1)+(b_3 - b_2)+ \ldots + (b_k - b_{k-1}) + (b_{k+1} - b_k) \\
         &= b_{k+1} - b_1
\end{align*}
where $(\Delta b)_n = b_{n+1} - b_n$ is the forward difference. Finding the partial sum has now been reduced to finding a sequence $b_n$ such that $(\Delta b)_n = a_n$.
We will find $b$, the antiderivative of $a$, using falling powers, which are defined by
$$
    n^{\underline{k}} = n(n-1)(n-2)\ldots (n-k+1)
$$
where $k$ is an integer and, by a second definition,
$n^{\underline{0}}=1$. For example
$$
    n^{\underline{3}} = n(n-1)(n-2).
$$
We now need one more result: the discrete derivative of $n^{\underline{k}}$ is given by
\begin{align*}
\Delta n^{\underline{k}} &= (n+1)^{\underline{k}} - n^{\underline{k}} \\
             &= (n+1)n^{\underline{k-1}} - n^{\underline{k-1}}(n-k+1) \\
             &= kn^{\underline{k-1}}
\end{align*}
Let's now find the partial sum for a particular case:
\begin{align}
\sum^{k}_{n=1}n(n+1)(n+2) &= \sum^{k}_{n=1} (n+2)^{\underline{3}} \\
&= \sum^{k}_{n=1} \Delta \left[\frac{1}{4} (n+2)^{\underline{4}}\right] \\
&= \frac{(k+3)(k+2)(k+1)k}{4} - \require{cancel}\cancelto{0}{\frac{(1+2)(1+1)(1-0)(1-1)}{4}}
\end{align}
The general case:
\begin{align}
\sum^{k}_{n=1} (n+p)^{\underline{p+1}} &= \sum^{k}_{n=1} \Delta \left[\frac{1}{p+2} (n+p)^{\underline{p+2}}\right] \\
&= \frac{(k+1+p)(k+p)\ldots [(k+1+p)-(p+2)+1)]}{p+2} \\
&= \frac{(k+1+p)(k+p)\ldots k}{p+2}
\end{align}
where $p>0$ is an integer.
