TITLE: Is $(7,4)$ the only non-trivial integer solution for $(n)_k=n!$?
QUESTION [6 upvotes]: I accidentally noticed that:
$$(7)_4=7 \cdot 8 \cdot 9 \cdot 10=2 \cdot  3 \cdot  4 \cdot  5 \cdot  6 \cdot  7=7!$$
Here  $(n)_k$ is the Pochhammer symbol.

I wonder, are there any other non-trivial integer solutions $(n,k)$?
$$(n)_k=n!$$

Among the ones I consider trivial we have $(0,0),(1,0),(1,1),(2,1)$. Somehow, I'm sure that I will get a lot of comments with these four solutions.
This is the implicit plot of the equivalent equation:
$$\Gamma (n+k)=n \Gamma^2 (n)$$

REPLY [3 votes]: If $n+k\geq 14$, then there are at two primes in the interval $\left(\frac{n+k-1}{2},n+k-1\right]$.  Therefore, $(n+k-1)!$ cannot equal $n\cdot \big((n-1)!\big)^2$.  This leaves the cases $n+k\leq 13$ to be dealt with.