Square root of four consecutive integers plus one

Compute $\sqrt{120\cdot121\cdot122\cdot123+1}$ exactly.

$14761$. Pairing outer and inner factors, $n(n+3)\cdot(n+1)(n+2)=(n^2+3n)(n^2+3n+2)$, so adding $1$ gives $(n^2+3n+1)^2$. With $n=120$ the root is $14400+360+1=14761$.