Finding a closed formula for $1\cdot2\cdot3\cdots k +\dots + n(n+1)(n+2)\cdots(k+n-1)$ Considering the following formulae:
(i) $1+2+3+..+n = n(n+1)/2$
(ii) $1\cdot2+2\cdot3+3\cdot4+...+n(n+1) = n(n+1)(n+2)/3$
(iii) $1\cdot2\cdot3+2\cdot3\cdot4+...+n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4$
Find and prove a 'closed formula' for the sum
$1\cdot2\cdot3\cdot...\cdot k + 2\cdot3\cdot4\cdot...\cdot(k+1) + ... + n(n+1)(n+2)\cdot...\cdot (k+n-1)$
generalizing the formulae above.
I have attempted to 'put' the first 3 formulae together but I am getting nowhere and wondered where to even start to finding a closed formula.
 A: If you divide both sides by $k!$ you will get binomial coefficients and you are in fact trying to prove
$$\binom kk + \binom{k+1}k + \dots + \binom{k+n-1}k = \binom{k+n}{k+1}.$$
This is precisely the identity from this question.
The same argument for $k=3$ was used here.

Or you can look at your problem the other way round: If you prove this result about finite sums
$$\sum_{j=1}^n j(j+1)\dots(j+k-1)= \frac{n(n+1)\dots{n+k-1}}{k+1},$$
you also get a proof of the identity about binomial coefficients.
A: For a fixed non-negative $k$, let $$f(i)=\frac{1}{k+1}i(i+1)\ldots(i+k).$$
Then $$f(i)-f(i-1)=i(i+1)\ldots(i+k-1).$$
By telescoping, 
$$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\sum_{i=1}^n\left(f(i)-f(i-1)\right)=f(n)-f(0)=f(n)$$
and we are done.
A: The pattern looks pretty clear: you have
$$\begin{align*}
&\sum_{i=1}^ni=\frac12n(n+1)\\
&\sum_{i=1}^ni(i+1)=\frac13n(n+1)(n+2)\\
&\sum_{i=1}^ni(i+1)(i+2)=\frac14n(n+1)(n+2)(n+3)\;,
\end{align*}\tag{1}$$
where the righthand sides are closed formulas for the lefthand sides. Now you want 
$$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\;;$$
what’s the obvious extension of the pattern of $(1)$? Once you write it down, the proof will be by induction on $n$.
Added: The general result, of which the three in $(1)$ are special cases, is $$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\frac1{k+1}n(n+1)(n+2)\dots(n+k)\;.\tag{2}$$ For $n=1$ this is $$k!=\frac1{k+1}(k+1)!\;,$$ which is certainly true. Now suppose that $(2)$ holds. Then
$$\begin{align*}\sum_{i=1}^{n+1}i(i+1)&(i+2)\dots(i+k-1)\\
&\overset{(1)}=(n+1)(n+2)\dots(n+k)+\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\\
&\overset{(2)}=(n+1)(n+2)\dots(n+k)+\frac1{k+1}n(n+1)(n+2)\dots(n+k)\\
&\overset{(3)}=\left(1+\frac{n}{k+1}\right)(n+1)(n+2)\dots(n+k)\\
&=\frac{n+k+1}{k+1}(n+1)(n+2)\dots(n+k)\\
&=\frac1{k+1}(n+1)(n+2)\dots(n+k)(n+k+1)\;,
\end{align*}$$
exactly what we wanted, giving us the induction step. Here $(1)$ is just separating the last term of the summation from the first $n$, $(2)$ is applying the induction hypothesis, $(3)$ is pulling out the common factor of $(n+1)(n+2)\dots(n+k)$, and the rest is just algebra.
A: I asked exactly this question a couple of days ago, here:
Telescoping series of form $\sum (n+1)\cdot...\cdot(n+k)$

My favourite solution path so far is 
to start with the hockey stick identity.
A: From (i), (ii) and (iii) it is reasonable to guess that your sum will be 
$$n(n+1)\cdot...\cdot(n+k)/(k+1)$$
Try to prove this by induction. 
