$(1 \cdot 2 \cdot ... \cdot m)^w+ (2 \cdot 3 \cdot ... \cdot (m+1))^w+...+(n \cdot(n+1)+ \cdot ... \cdot (n+m-1))^w=?$ I started to read some book on elementary number theory and preliminary chapter asks to establish some formulas by mathematical induction:
There is this formula:
$$1+2+...+n= \dfrac {n(n+1)}{2}$$
And this one:
$$ 1\cdot 2 + 2\cdot 3 + ... + n\cdot (n+1)= \dfrac {n(n+1)(n+2)}{3}$$
With some pencil-and-paper work I established that this should hold:
$$1\cdot 2 \cdot ... \cdot m + 2\cdot 3 \cdot ... \cdot (m+1) + ... + n \cdot (n+1) \cdot ... \cdot (n+m-1)= \dfrac {n(n+1)...(n+m-1)(n+m)}{m+1}$$
I did not prove this formula that I established but just checked some cases and it seems to hold.
It can be written in the form of a hockey-stick identity, I think, so it holds.
Now, I know about generalization of first formula that goes like this (Faulhaber´s formula):
$$1^w + 2^w + ... + n^w=\dfrac {1}{w+1} \cdot \displaystyle \sum_{j=0}^{w} { {w+1} \choose j} B_j n^{w+1-j}$$
How do the generalization $$(1 \cdot 2 \cdot ... \cdot m)^w+ (2 \cdot 3 \cdot ... \cdot (m+1))^w+...+(n \cdot(n+1)+ \cdot ... \cdot (n+m-1))^w=?$$ look like? That is, what is on the right side?
 A: I think there is no simple closed form. Using your notation, if we denote
$$w=a_{1}$$
$$\sum\limits_{q=1}^{m} a_{q}=b_{m}$$
so
$$\sum\limits_{k=1}^{n} \left(\frac{m+k-1}{k-1}\right)^{w}=\sum\limits_{s=1}^{m-1}\sum\limits_{a_{s+1}=0}^{a_{s}}\frac{a_{1}!}{a_{m}!}\prod\limits_{t=1}^{m-1}\frac{1}{(a_{t}-a_{t+1})!}\left(\left[{a_{1}}\atop {a_{m-t+1}}\right]\right)^{a_{t}-a_{t+1}}\sum\limits_{k=1}^{n}k^{b_{m}}$$
which is a little bit ugly.
A: For your first conjecture:
Write
$1\cdot 2 \cdot ... \cdot m + 2\cdot 3 \cdot ... \cdot (m+1) + ... + n \cdot (n+1) \cdot ... \cdot (n+m-1)= \dfrac {n(n+1)...(n+m-1)(n+m)}{m+1}
$
in the form
$s(n, m)
=t(n, m)
$
where
$s(n, m)
=\sum_{k=1}^n \prod_{j=k}^{k+m-1} j
$
and
$t(n, m)
=\dfrac{\prod_{j=n}^{n+m}j}{m+1}
$.
I will prove this 
by induction on $n$.
For $n=1$
this is
$=\prod_{j=1}^{m} j
=\dfrac{\prod_{j=1}^{1+m}j}{m+1}
$
or
$m! = \dfrac{(m+1)!}{m+1}
$
so that
$s(1, m) = t(1, m)$.
Consider
$\begin{array}\\
s(n+1,m)-s(n,m)
&=\sum_{k=1}^{n+1} \prod_{j=k}^{k+m-1} j-\sum_{k=1}^n \prod_{j=k}^{k+m-1} j\\
&=\prod_{j=n+1}^{n+m} j\\
\end{array}
$
and
$\begin{array}\\
t(n+1, m)-t(n, m)
&=\dfrac{\prod_{j=n+1}^{n+1+m}j}{m+1}-\dfrac{\prod_{j=n}^{n+m}j}{m+1}\\
&=\dfrac{\prod_{j=n+1}^{n+1+m}j-\prod_{j=n}^{n+m}j}{m+1}\\
&=\dfrac{\prod_{j=n+1}^{n+m}j\left((n+1+m)-n\right)}{m+1}\\
&=\dfrac{\prod_{j=n+1}^{n+m}j\left(m+1\right)}{m+1}\\
&=\prod_{j=n+1}^{n+m}j\\
&=s(n+1,m)-s(n,m)\\
\end{array}
$
Since
$s(1, m) = t(1, m)$,
this shows that
$s(n, m) = t(n, m)$.
