formula for $\left(1\cdot2\cdot...\cdot k\right)+...+\left(n\left(n+1\right)...\left(n+k- 1\right)\right)$ proof if $k$ is a constant then for every $n$ natural numbers we have:

$$\left(1\cdot2\cdot...\cdot
k\right)+\left(2\cdot3\cdot...\cdot\left(k+1\right)\right)+...+\left(n\left(n+1\right)...\left(n+k-
1\right)\right)=\frac{n\left(n+1\right)\left(n+2\right)...\left(n+k\right)}{k+1}.$$
  clearly $$\left(1\cdot2\cdot...\cdot 
k\right)+\left(2\cdot3\cdot...\cdot\left(k+1\right)\right)+...+\left(n\left(n+1\right)...\left(n+k-
1\right)\right)=\sum_{m=1}^{n}\left(\prod_{j=0}^{k-1}\left(m+j\right)\right)$$
$$=\sum_{m=1}^{n}\left(\Gamma\left(m+j+1\right)\right)$$where$\left(0\le j\le k-1\right)$ and 
  $\Gamma(x)$ is Gamma function,
  or equivalently $$=\sum_{m=1}^{n}\int_{0}^{∞}x^{\left(m+j\right)}e^{-x}dx$$
  but I could not find the given formula.

 A: Note that by the Hockey-stick identity we have that
$$1\cdot2\cdots k+2\cdot3\cdots(k+1)+\dots+n\cdot(n+1)\cdots(n+k-1)$$
$$\begin{align}
&=k!\binom{k}{k}+k!\binom{k+1}{k}+\dots+k!\binom{n+k-1}{k}\\
&=k!\sum_{i=k}^{n+k-1}\binom{i}{k}\\
&=k!\binom{n+k}{k+1}\\
&=k!\cdot\frac{n\cdot(n+1)\cdots(n+k)}{(k+1)!}\\
&=\frac{n\cdot(n+1)\cdots(n+k)}{k+1}\\
\end{align}$$
A: We can prove this by induction. The base case here is when $n=1.$ Clearly,
$n\cdot(n+1)\cdot(n+2)\cdot\dots\cdot(n+k-1)=1\cdot(2)\cdot(3)\cdot\dots\cdot(k)$ 
and 
$\dfrac{(n)\cdot(n+1)\cdot(n+2)\cdot\dots\cdot(n+k)}{1+k}=(1)\cdot(2)\cdot(3)\cdot\dots\cdot(k)=n\cdot(n+1)\cdot(n+2)\cdot\dots\cdot(n+k-1)$ 
so the base case holds. Assume the statement is true for some $y\in\mathbb{N}.$ Then 
$$
\begin{align}(1)\cdot(2)\cdot\dots\cdot(k)+(2)\cdot(3)\cdot\dots\cdot(k+1)+\dots+(y+1)\cdot(y+2)\cdot\dots\cdot(y+k)=[(1)\cdot(2)\cdot\dots\cdot(k)+(2)\cdot(3)\cdot\dots\cdot(k+1)\\+(y)\cdot(y+1)\cdot\dots\cdot(y+k-1)]+(y+1)\cdot(y+2)\cdot\dots\cdot(y+k)\\=\dfrac{(y)\cdot(y+1)\cdot(y+2)\cdot\dots\cdot(y+k)}{1+k}+(y+1)\cdot(y+2)\cdot\dots\cdot(y+k)=\dfrac{(y+1)\cdot(y+2)\dots(y+k+1)}{k+1},\end{align}
$$ 
so the equation holds by induction.
