Infinite Series $\sum\limits_{n=1}^{\infty}\frac{1}{\prod\limits_{k=1}^{m}(n+k)}$

$$\sum_{n=1}^{\infty}\frac{1}{\prod_{k=1}^{m}(n+k)}= \sum_{n=1}^{\infty}\frac{n!}{(n+m)!} = \frac{1}{m!}\sum_{n=1}^{\infty} \frac{1}{\binom{m+n}{n}}$$

Now

$$\frac{1}{\binom{m+n}{n}} = m \int_0^1 dx \, x^{m-1} (1-x)^n$$

so that the sum in question is equal to, upon reversing sum and integral

$$\begin{align}m \int_0^1 dx \, x^{m-1} \sum_{n=1}^{\infty} (1-x)^n &= m \int_0^1 dx \, x^{m-1} \frac{1-x}{x} \\ &= m \int_0^1 dx \, \left ( x^{m-2}-x^{m-1}\right) \\ &= m \left (\frac{1}{m-1}-\frac{1}{m}\right ) \\ &= \frac{1}{m-1} \end{align}$$

Putting this all together,

$$\sum_{n=1}^{\infty}\frac{1}{\prod_{k=1}^{m}(n+k)}= \frac{1}{(m-1) m!}$$

as was to be shown.

Note that $\prod\limits_{k=1}^{m}(n+k) = \frac{(n+m)!}{n!}$, then prove by induction the explicit formula for the $s$'th partial sum $$ \sum_{n=1}^{s}\frac{1}{\prod_{k=1}^{m}(n+k)} = \sum_{n=1}^{s}\frac{n!}{(n+m)!} = \frac{1}{m-1} \left( \frac{1}{m!} - \frac{(s+1)!}{(s+m)!}\right) $$ and take the limit $s \rightarrow \infty.$

Another approach without using hidden beta and gamma functions where one gets to excercise manipulation of binomial coefficients and generating functions is to use $$\prod_{k=1}^m \frac{1}{n+k} = \frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{n+k+1}.$$ Now compute how often the fraction $1/q$ where $q\ge m+1$ occurs on the right side and with what coefficients. It occurrs for the first time in the term for $n=q-m$ and for the last time in $n=q-1,$ so its coefficient is (keeping in mind that $k = q-n-1$) $$\sum_{n=q-m}^{q-1} (-1)^{q-n-1} \binom{m-1}{q-n-1}.$$ Put $p=q-n-1$ in the above sum to get $$\sum_{p=q-(q-m)-1}^{q-(q-1)-1} (-1)^p \binom{m-1}{p} = \sum_{p=m-1}^0 (-1)^p \binom{m-1}{p} = \sum_{p=0}^{m-1} (-1)^p \binom{m-1}{p} = 0,$$ so for $q\ge m+1$ the sum telescopes and the contribution of the value $1/q$ is zero.

The leftover contribution is $$\sum_{q=2}^m \frac{1}{q}\sum_{n=1}^{q-1} (-1)^{q-n-1} \binom{m-1}{q-n-1} = \sum_{q=2}^m \frac{1}{q}\sum_{n=0}^{q-2} (-1)^n \binom{m-1}{n}$$

To conclude we ask where $\binom{m-1}{n}$ appears and for what values of $q$, getting $$ \sum_{n=0}^{m-2} (-1)^n \binom{m-1}{n} \sum_{q=2+n}^m \frac{1}{q} = \sum_{n=0}^{m-2} (-1)^n \binom{m-1}{n} \left(H_m - H_{n+1}\right).$$ Split the sum in two to obtain first, $$H_m \sum_{n=0}^{m-2} (-1)^n \binom{m-1}{n} = H_m (0 - (-1)^{m-1} ) = (-1)^m H_m$$ and second, $$ \sum_{n=0}^{m-2} (-1)^n \binom{m-1}{n} H_{n+1} = \sum_{n=0}^{m-1} (-1)^n \binom{m-1}{n} H_{n+1} - (-1)^{m-1} H_m = \sum_{n=0}^{m-1} (-1)^n \binom{m-1}{n} H_{n+1} + (-1)^m H_m$$ so the difference between the two is $$- \sum_{n=0}^{m-1} (-1)^n \binom{m-1}{n} H_{n+1}.$$ This is the binomial transform of $H_{n+1}$ and since $$\sum_{n\ge 1} H_{n+1} z^n = f(z) = \frac{1}{z} \frac{1}{1-z} \log \frac{1}{1-z}$$ the sum has generating function $$\frac{1}{1-z} f\left(\frac{z}{z-1}\right) = \frac{1}{z} (1-z) \log \frac{1}{1-z} = \left(\frac{1}{z}-1\right) \log \frac{1}{1-z} \\= \sum_{m\ge 1} \left(\frac{1}{m+1}-\frac{1}{m}\right) z^m = -\sum_{m\ge 1} \frac{1}{m(m+1)} z^m.$$ Observe that we need the coefficient of $z^{m-1}.$ It follows that the original sum has value $$ \frac{1}{(m-1)!} \frac{1}{(m-1)m} = \frac{1}{(m-1)m!}$$ as was to be shown.

I will prove the partial fraction identity in an additional answer.