\(\displaystyle \sum_{k=1}^{\infty}\frac{1}{\prod_{j = 0}^p (k + j)}= \sum_{k=1}^{\infty}\frac{\Gamma(k)}{\Gamma(p+k+1)}\text{ (Rewrite in terms of Gamma function)}\)
\(\displaystyle =\frac{1}{\Gamma(p+1)}\sum_{k=1}^{\infty}\frac{\Gamma(k)\Gamma(p+1)}{\Gamma(p+k+1)} \text{ (Complete the Beta function)}\)
\(\displaystyle =\frac{1}{\Gamma(p+1)} \sum_{k=1}^{\infty}\Beta(p+1,k)\)
\(\displaystyle =\frac{1}{\Gamma(p+1)} \sum_{k=1}^{\infty} \int_{0}^{1} \alpha^{k-1}(1-\alpha)^{p}\, d\alpha\)
\(\displaystyle =\frac{1}{\Gamma(p+1)}\int_{0}^{1}\sum_{k=1}^{\infty} \alpha^{k-1}(1-\alpha)^{p}\, d\alpha\)
\(\displaystyle =\frac{1}{\Gamma(p+1)}\int_{0}^{1}(1-\alpha)^{p-1}\, d\alpha\)
\(\displaystyle =\frac{1}{\Gamma(p+1)}\cdot \frac{1}{p}= \frac{1}{p!\times p}\)
where [imath]\Gamma()[/imath] is the Gamma function and [imath]\Beta()[/imath] is the Beta function.