Simpler Way to Write a sum of factorial I have a fairly simple question. I was wondering if there is a simpler way to write the following: $$\sum_{k=0}^{x-1}\frac{(n+k+1)!}{k!}$$
So for example, given $n=4$ and $x=3$, we would have $$1\cdot 2\cdot 3\cdot 4\cdot 5+2\cdot 3\cdot 4\cdot 5\cdot 6+3\cdot 4\cdot 5\cdot 6\cdot 7=3360$$
Is there a simpler way to write this or do this calculation?
Edit: Ok I have simplified the sum above to the following: $$\frac{x(n+x+1)!}{(n+2)(x!)}$$
Can I simplify this further?
 A: Well, if we are to believe WolframAlpha, then yes there is: 
$$\sum_{k=1}^x\frac{(n+k+1)!}{k!}= \frac{(x+1)(n+x+2)!-(n+2)!(x+1)!}{(n+2)(x+1)!}$$
Expect an edit of me trying to prove that... 
A: $$\sum_{k=0}^{x-1} \dfrac{(n+k+1)!}{k!} = \dfrac{x (n+x+1)!}{(n+2) x!} $$
A: Note that we can write 
$$\frac{(n+k+1)!}{k!}=\frac{1}{n+2}\left(\frac{(k+n+2)!}{k!}-\frac{(k+n+1)!}{(k-1)!}\right)$$
Therefore, evaluating the telescoping sum is trivial and yields
$$\begin{align}
\sum_{k=0}^{x-1}\frac{(n+k+1)!}{k!}&=(n+1)!+\sum_{k=1}^{x-1}\frac{(n+k+1)!}{k!}\\\\
&=(n+1)!+\frac{1}{n+2}\left(\frac{(x+n+1)!}{(x-1)!}-\frac{(n+2)!}{1!}\right)\\\\
&=\frac{1}{n+2}\frac{(x+n+1)!}{(x-1)!}\\\\
&=\frac{1}{n+2}\prod_{\ell =0}^{n+1}(x+\ell)\\\\
&=(n+1)!\binom{x+n+1}{x-1}
\end{align}$$
A: Another way, but not sure it is simpler:
$$
\sum_{k=1}^x \frac{(n+k+1)!}{k!} = \sum_{k=1}^x \binom{n+k+1}{k} (n+1)!
$$
A: Mostly a commentary to the answers of Αδριανός and Robert Israel: using Gosper's algorithm (AntiDifference((n+k+1)!/k!,k) in Maxima):
$$
\frac{(n+k+1)!}{k!} = 
{{(n+k+2)!}\over{k!(n+2)}} -
{{(n+k+1)!}\over{(k-1)!(n+2)}}
$$and the sum telescopes.
