Computing $\sum_{n=1}^∞\frac{1}{(n+1)(n+2)(n+3)....(n+p)}$

I have a series question which I can not solve for two days.

$$\sum_{n=1}^∞\frac{1}{(n+1)(n+2)(n+3)....(n+p)} = \frac{1}{p.p!}$$

How can I prove and solve this equation?

Note the following $$\sum_{n=1}^∞\frac{1}{(n+1)(n+2)(n+3)....(n+p)}$$

Using partial fraction decomposition the summation becomes $$\frac{1}{p-1}\sum_{n=1}^∞ \left(\frac{1}{(n+1)(n+2)....(n+p-1)}-\frac{1}{(n+2)(n+3)....(n+p)}\right)$$ We can use the telescoping series. The consecutive terms cancel. So your answer is incorrect. The correct answer would be $$\frac{1}{p-1} \times \frac{1}{(1+1)(1+2)....(1+p-1)}=\frac{1}{(p-1)p!}$$ We are done! As mentioned by @ThomasAndrews, the formula would be correct if the denominator was $n(n+1)(n+2)(n+3)....(n+p)$ which can be confirmed through the same method.

$$\frac{1}{(n+1)(n+2)(n+3)....(n+p)} = \frac{n!}{(n+p)!}=\frac{1}{\Gamma(p)}\frac{\Gamma(n+1)\Gamma(p)}{\Gamma(n+p+1)}=\frac{1}{\Gamma(p)}\beta(n+1,p)$$ then \begin{eqnarray} \sum_{n=1}^\infty\frac{1}{(n+1)(n+2)(n+3)....(n+p)} &=& \sum_{n=1}^\infty\frac{1}{\Gamma(p)}\beta(n+1,p)\\ &=& \frac{1}{\Gamma(p)}\sum_{n=1}^\infty\int_0^1t^{n}(1-t)^{p-1}dt\\ &=& \frac{1}{\Gamma(p)}\int_0^1\sum_{n=1}^\infty t^{n}(1-t)^{p-1}dt\\ &=& \frac{1}{\Gamma(p)}\int_0^1\frac{t}{1-t}(1-t)^{p-1}dt\\ &=& \frac{1}{\Gamma(p)}\int_0^1t(1-t)^{p-2}dt\\ &=& \frac{1}{\Gamma(p)}\int_0^1t^{p-2}(1-t)dt\\ &=& \frac{1}{\Gamma(p)}\int_0^1t^{p-2}-t^{p-1}dt\\ &=& \frac{1}{\Gamma(p)}\frac{1}{p(p-1)}\\ &=& \frac{1}{p!(p-1)} \end{eqnarray} For $p=2$ we know $$\sum_{n=1}^\infty\frac{1}{(n+1)(n+2)}=\sum_{n=1}^\infty\frac{1}{n+1}-\frac{1}{n+2}=\frac12$$ that shows our answer is true.