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.

The simplest way is a computer program.  In matlab: 
x=1;
for n=1:51
x = mod(x*n,101);
end
x
