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?
 A: 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.
A: The easy way to deal, for example, with $\sum_{i=1}^n i(i+1)(i+2)(i+3)$ is to let $F(i)=i(i+1)(i+2)(i+3)(i+4)$. We calculate $F(i)-F(i-1)$. We get
$$i(i+1)(i+2)(i+3)(i+4)-(i-1)(i)(i+1)(i+2)(i+3).$$
There is a common factor of $i(i+1)(i+2)(i+3)$. When we "take it out" we are left with $(i+4)-(i-1)=5$. 
Let $G(i)=\frac{F(i)}{5}$. Then by our calculation $i(i+1)(i+2)(i+3)=G(i)-G(i-1)$.
Now consider the sum $\sum_{i=1}^n i(i+1)(i+2)(i+3)$. This is 
$$(G(1)-G(0))+(G(2)-G(1))+G(3)-G(2)) +\cdots+(G(n)-G(n-1)).$$
Observe the telescoping. Since $G(0)=0$, the above sum is equal to $G(n)$. Thus
$$\sum_{i=1}^n i(i+1)(i+2)(i+3)=G(n)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}.$$
Exactly the same idea works in general.  
A: The hypothesized equality can be written as follows: for any $m$, we conjecture
$$
\sum^{k}_{n=1}\frac{(n+m)!}{(n-1)!}=\frac{(k+m+1)!}{(m+2)(k-1)!}
$$
Dividing both sides by $(m+1)!$, we have
$$
\sum^{k}_{n=1}\frac{(n+m)!}{(n-1)!(m+1)!}=\frac{(k+m+1)!}{(m+2)!(k-1)!}
$$
Or, in other words
$$
\sum^{k}_{n=1}\binom{n+m}{m+1}=\binom{k+m+1}{m+2}
$$
I'm not sure how to prove this (yet), but it seems very likely that there's a neat trick for all this.
A: You can argue any given case by induction.   I will take your last,$$\sum^{k}_{n=1}n(n+1)(n+2)=\frac{k(k+1)(k+2)(k+3)}{4}$$ for the example, but I think it is easy to see how it gets carried forward.  The base case is simply $1\cdot 2\cdot 3=1\cdot 2\cdot 3\cdot \frac 44$  If it is true up to $k$, then $$\sum^{k+1}_{n=1}n(n+1)(n+2)\\=\sum^{k}_{n=1}n(n+1)(n+2)+(k+1)(k+2)(k+3)\\=\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)\frac {k+4-k}4\\=\frac{(k+1)(k+2)(k+3)(k+4)}{4}$$
A: [I haven't finished yet. Will be gradually improved]
If we do it this way, then probably, it will become more insightful.
$$
\sum^{k}_{n=1}1=k
$$
then we preserve everything from the right side exactly as it is.
$$
\sum^{k}_{n=1}n=\frac{k(k+1)}{1 \cdot 2} 
$$
The following picture explains this sum:

Now again, preserve everything from the right side:
$$
\sum^{k}_{n=1}\frac{n(n+1)}{1 \cdot 2}=\frac{k(k+1)(k+2)}{1 \cdot 2 \cdot 3} 
$$
[there will be a picture of the 6 pyramids composed into rectangular parallelepiped]
$$
\sum^{k}_{n=1}\frac{k(k+1)(k+2)}{1 \cdot 2 \cdot 3}=\frac{k(k+1)(k+2)(k + 3)}{1 \cdot 2 \cdot 3 \cdot 4}
$$
[There should be a description of the connection between simple combinations and combinations with repetition]
$$
{n \choose k} = \frac{n \cdot (n - 1) \cdot \ldots \cdot (n - k + 1)}{k \cdot (k - 1) \cdot \ldots \cdot 1} = \frac{n^{\underline{k}}}{k!}
$$
$$
\left(\!\middle(\genfrac{}{}{0pt}{}{n}{k}\middle)\!\right) = \frac{n \cdot (n + 1) \cdot \ldots \cdot (n + k - 1)}{k \cdot (k - 1) \cdot \ldots \cdot 1} = \frac{n^{\overline{k}}}{k!}
$$
