If $m,n\in\Bbb{Z}_+$ s.t. $mQuestion:

If $m,n\in\Bbb{Z}_+$ s.t. $m<n$, is $$x_{n,m}=\frac{\Gamma(n+1)\Gamma(n+2)}{\Gamma(m+1)\Gamma(m+2)\Gamma(n-m+1)\Gamma(n-m+2)}$$ an integer? ($\Gamma$ is the gamma function, the extension of the factorial function)

My attempt:
We have
\begin{align}x_{n,m}=\frac{\Gamma(n+1)\Gamma(n+2)}{\Gamma(m+1)\Gamma(m+2)\Gamma(n-m+1)\Gamma(n-m+2)}\\
=\frac{n!(n+1)!}{m!(m+1)!(n-m)!(n-m+1)!}\\
=\frac{\prod_{j=1}^m(n-j+1)(n-j+2)}{m!(m+1)!}
\end{align}
I explored some basic cases:

*

*When $m=1$, $x_{n,1}=\frac{n(n+1)}{2}$, which is clearly an integer for any $n>m$ because either $n$ or $n+1$ is even.

*When $m=2$, $x_{n,2}=\frac{n^2(n+1)(n-1)}{12}$, which is still an integer for any $n>m$ because one of $n,(n+1), (n-1)$ is divisible by $3$, and two of them must be divisible by $2$, since if $2|n$, then $4|n^2$, else if $2$ is not a factor of $n$, then both $n+1$ and $n-1$ is divisible by $2$. So the whole thing is divisible by $4$ also.

*When $m=3$, the denominator becomes $144$, I could also brutally prove that it's still an integer, but more importantly it seems to suggest that $x_{n,m}$ is always an integer, so I'm wondering if this could be rigorously proved by induction.

Induction Process:
The base case is already verified in the preceding section, so suppose $x_{n,m}\in\Bbb Z$ for $m=1,...,m-1$, then for $p=1,...,m$
\begin{align}
x_{n+1,p}=\frac{\prod_{j=1}^{p}(n+1-j+1)(n+1-j+2)}{p!(p+1)!}\\
=\frac{\prod_{j=1}^m(n+1-j+1)(n+1-j+2)}{m!(m+1)!}\frac{\prod_{j=m+1}^{p}(n+1-j+1)(n+1-j+2)}{\prod_{j=m+1}^pj(j+1)}\\
\end{align}
and this doesn't tell me anything except that it somehow looks like the proof of Pascal's identity... The proof of my claim could be elementary but I didn't see any plausible argument at this point...
Thanks in advance.
 A: Starting from the middle line of your attempt gives
$$\begin{equation}\begin{aligned}
& \frac{n!(n+1)!}{m!(m+1)!(n-m)!(n-m+1)!} \\
& = \left(\frac{n!}{(m+1)!(n-m)!}\right)\left(\frac{(n+1)!}{m!(n-m+1)!}\right) \\
& = \frac{1}{n+1}\left(\frac{(n+1)!}{(m+1)!(n-m)!}\right)\left(\frac{(n+1)!}{m!(n-m+1)!}\right) \\
& = \frac{1}{n+1}\binom{n+1}{m+1}\binom{n+1}{m}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
To prove \eqref{eq1A} is an integer requires
$$(n + 1) \; \left| \; \binom{n+1}{m+1}\binom{n+1}{m} \right. \tag{2}\label{eq2A}$$
Note almost the exact same thing was asked, and proven in quite different ways in  $2$ answers, in Prove using combinatorics $n \mid \binom{n}{m} \binom{n}{m-1}$, with their $n$ and $m$ being replaced by $n + 1$ and $m + 1$, respectively, in your case. For example, the answer there given by Batominovski is:

Maybe this is not a combinatorial proof, but $$\frac{1}{n}\,\binom{n}{m}\,\binom{n}{m-1}=\binom{n-1}{m-1}\,\binom{n+1}{m}-\binom{n}{m}\,\binom{n}{m-1}\,.$$
This is because
$$\frac{1}{n}\,\binom{n}{m}\,\binom{n}{m-1}=\frac{1}{m}\,\binom{n-1}{m-1}\,\binom{n}{m-1}=\frac{1}{n+1}\,\binom{n-1}{m-1}\,\binom{n+1}{m}\,.$$
The above result is due to the identity $\displaystyle\binom{k}{r}=\frac{k}{r}\,\binom{k-1}{r-1}$, which has a combinatorial proof.

Adjusting the identity in the first equation above to your use of $n$ and $m$ gives, on its right side,
$$\begin{equation}\begin{aligned}
& \binom{n}{m}\,\binom{n+2}{m+1} - \binom{n+1}{m+1}\,\binom{n+1}{m} \\
& = \left(\frac{n!}{m!(n-m)!}\right)\left(\frac{(n+2)!}{(m+1)!(n-m+1)!}\right) - \\
& \; \; \; \left(\frac{(n+1)!}{(m+1)!(n-m)!}\right)\left(\frac{(n+1)!}{m!(n-m+1)!}\right) \\
& = \left(\frac{n!(n+1)!}{m!(m+1)!(n-m)!(n-m+1)!}\right)\left((n+2) - (n+1)\right) \\
& = \frac{n!(n+1)!}{m!(m+1)!(n-m)!(n-m+1)!}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
which is the first line of \eqref{eq1A}. Since binomial coefficients are always integers, this confirms your relation is always an integer.
