Has anyone extended Wilson's Theorem?

ok here's something little i derived (p-3)! = (p-1)/2 mod p so applying it putting p=9900000959 in it we get 4950000479

We know by Wilson's Theorem that for any prime $p$ ,

$(p-1)! \equiv$ $-1$ $mod$ $p$

Since $9900000959$ is prime, we have

$(9900000959-1)! \equiv$ $-1$ $mod$ $9900000959$ or $9900000958! \equiv$ $9900000958$ $mod$ $p$

It can be observed that $9900000958! = 9900000958 \times 9900000957 \times 9900000956!$

Also, $9900000958 \equiv$ $-1$ $mod$ $9900000959$ and $9900000957 \equiv$ $-2$ $mod$ $9900000959$

So,

$9900000958 \times 9900000957 \times 9900000956! \equiv$ $(-1)\times(-2)\times 9900000956!\ \equiv$ $9900000958$ $mod$ $9900000959$

Thus,

$9900000956! \equiv$ $\frac{9900000958}{(-1)\times(-2)}\equiv$ $4950000479$ $mod$ $9900000959$

From wilsons theorem it follows that, the answer is $\frac{-1}{2}$ modulo that prime we can easliy find the modular inverse of 2 here.Simply, $\frac{prime+1}{2}$ for that minus sign we subtract the modular inverse of 2 from that prime which results in 4950000479