How to calculate $51! \ mod \ 101$? I tried almost everything I know
Even tried to calculate it from Wilson's theorem and what I got was
$$(101-50)! \equiv 51! \equiv (101 + 49!)^{-1} mod \ 101$$
I derived it from 
$$(p-1)! \equiv -1 \ mod \ p$$
$$(p-2)! \equiv 1 \ mod \ p$$
$$(p-3)! \equiv (p-2)^{-1} \ mod \ p$$
$$(p-4)! \equiv (p+6)^{-1} \ mod \ p$$
and so on...
but calculating right side is as hard as calculating left side and I can't just do it with my pen and paper. Am I going in right direction? I would appreciate some help on this.

We already know that $100! \equiv -1$ (mod $101$).
Moreover, we have $i \equiv -(101-i)$ (mod $101$), for $1\leq i\leq 50$. Then, one has $50! \equiv (-1)^{50} 51\times 52\times \dots \times 100$, it implies that $50! \equiv 10$ (mod $101$) or $50! \equiv 91$ (mod $101$). But $50! \not\equiv 10$ (mod $101$) (Someone helps me?). So, we get $51! \equiv 91\times 51 \equiv -5$ (mod $101$).
