Summation of product of integers with consecutive integers as first $m$ factors It is well known that 
$$\sum_{r=1}^n r^\overline{m}=\frac {n^{\overline{m+1}}}{m+1}\\
\text{i.e.}
\sum_{r=1}^n \scriptsize(r+1)(r+2)\cdots (r+m-1)=\frac{n(n+1)(n+2)\cdots (n+m)}{m+1}\\
\text{which can also be written as }\\
\qquad \scriptsize m!\sum_{r=1}^n\binom {r+m-1}m=\frac 1{m+1}\binom {n+m}{m+1}$$.

Can it be shown that
  $$\sum_{r=1}^n \left[r^\overline{m}\prod_{k=1}^q (r+p_k)\right]
=\frac {n^{\overline{m+1}}}{m+1}P_q(n)\\
\text{i.e. }\tiny \sum_{r=1}^n r(r+1)(r+2)\cdots (r+m-1)(r+p_1)(r+p_2)\cdots (r+p_q)
=\frac{n(n+1)(n+2)\cdots (n+m)}{m+1}\cdot \left(a_qn^q+a_{q-1}n^{q-1}+\cdots+a_1 q+a_0\right)$$
  where $p_k$ are positive integers greater than $m$, and $P_q(n)$ is a polynomial of degree $q$ in $n$?

A few examples (with solutions from Wolframalpha):  
1. $\scriptsize\displaystyle\sum_{r=1}^n \boxed{r(r+1)}(r+5)=\frac 1{4}\boxed{n(n+1)(n+2)}(n+7)$  
2. $\scriptsize\displaystyle\sum_{r=1}^n \boxed{r(r+1)(r+2)}(r+4)(r+8)=\frac 1{12}\boxed{n(n+1)(n+2)(n+3)}(2n^2+30n+103)$  
3.$\scriptsize\displaystyle\sum_{r=1}^n \boxed{r(r+1)(r+2)(r+3)}(r+5)(r+8)=\frac 1{105}\boxed{n(n+1)(n+2)(n+3)(n+4)}(15n^2+235n+884)$  
 A: 
The assumption is correct (and in fact can be generalized) since we have
  \begin{align*}
\sum_{r=1}^nr^{\overline{m+1}}&=\frac{n^{\overline{m+2}}}{m+2}\\
&=\frac{n^{\overline{m+1}}}{m+1}\cdot\frac{m+1}{m+2}(n+m+1)\\
&=\frac{m+1}{m+2}(n+m+1)\sum_{r=1}^n r^{\overline{m}}\tag{1}
\end{align*}
  as well as
  \begin{align*}
\sum_{r=1}^nr^{\overline{m}} r&=\sum_{r=1}^nr^{\overline{m}}(r+m-m)\\
&=\sum_{r=1}^nr^{\overline{m+1}}-m\sum_{r=1}^nr^{\overline{m}}\tag{2}\\
\end{align*}
  and for positive integer $k$
  \begin{align*}
\sum_{r=1}^nr^{\overline{m}} r^k&=\sum_{r=1}^nr^{\overline{m}}(r+m-m)r^{k-1}\\
&=\sum_{r=1}^nr^{\overline{m+1}}r^{k-1}-m\sum_{r=1}^nr^{\overline{m}}r^{k-1}\\
\end{align*}

Let's look at OP's first example with 
\begin{align*}
\sum_{r=1}^nr(r+1)&=\sum_{r=1}^nr^{\overline{2}}=\frac{1}{3}n(n+1)(n+2)\tag{3}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{r=1}^nr^{\overline{2}}(r+5)}
&=\sum_{r=1}^nr^\overline{2}r+5\sum_{r=1}^nr^{\overline{2}}\\
&=\left(\sum_{r=1}^nr^{\overline{3}}-2\sum_{r=1}^nr^{\overline{2}}\right)+5\sum_{r=1}^nr^{\overline{2}}\tag{apply 2}\\
&=\frac{3}{4}(n+3)\sum_{r=1}^nr^{\overline{2}}+3\sum_{r=1}^nr^{\overline{2}}\tag{apply 1}\\
&=\frac{3}{4}(n+7)\sum_{r=1}^nr^{\overline{2}}\\
&\,\,\color{blue}{=\frac{1}{4}n(n+1)(n+2)(n+7)}\tag{apply 3}
\end{align*}
