We have that $719$ is a prime number. So by Wilson's theorem $718!+1\equiv 0\pmod {719}$ But $718!\equiv 712!\cdot 6!\pmod {719}$ and $6!=720\equiv 1\pmod {719}.$ So $719|712!+1.$

Are there any alternative solutions?

Note that $719$ is prime, and $718! \equiv -1 \pmod {719}$ by Wilson's. But $6! \equiv 1 \pmod {719}$, thus $719 \mid 712!+1$.

Notice that $713=23 \times 31$ is not a prime The smallest prime greater than $712$ is $719$ $\therefore 718!+1\equiv 0\ (mod\ 719)$ Besides, $718!+1\equiv 712! \cdot 713 \cdot 714 \cdot ......\cdot 718+1 \equiv 712! \cdot (-6) \cdot (-5) \cdot ......\cdot (-1)+1\equiv 712!\cdot 720+1 \equiv 712!\cdot 1+1\equiv 712!+1\ (mod\ 719)$ $\therefore 712!+1 \equiv 0\ (mod\ 719) \longrightarrow712!+1$ is NOT a prime