What is the most efficient way to compute factorials modulo a prime?

Do you know any algorithm that calculates the factorial after modulus efficiently?

For example, I want to program:

for(i=0; i<5; i++)
  sum += factorial(p-i) % p;

But, p is a big number (prime) for applying factorial directly $(p \leq 10^ 8)$.

In Python, this task is really easy, but i really want to know how to optimize.

(This answer was initially posted by the asker jonaprieto inside the question.)

I remember Wilson's theorem, and I noticed little things:

In the above program, it is better if I write:
$$\begin{align}
(p-1)! &\equiv -1 &\pmod p\\
(p-2)! &\equiv (p-1)! (p-1)^ {-1} \equiv \bf{1} &\pmod p\\
(p-3)! &\equiv (p-2)! (p-2)^ {-1} \equiv \bf{(p-2)^{-1}} &\pmod p\\
(p-4)! &\equiv (p-3)! (p-3)^ {-1} \equiv \bf{(p-2)^{-1}} \bf{(p-3)^{-1}} &\pmod p\\
\ (p-5)! &\equiv (p-4)! (p-4)^ {-1} \equiv \bf{(p-2)^{-1}} \bf{(p-3)^{-1}}\bf{(p-4)^{-1}} &\pmod p\\
\end{align}$$

And you can find $(p-i)^{-1}$ because $\operatorname{gcd}(p, p-i) = 1$, so with the extended Euclidian algorithm you can find the value of $(p-i)^{-1}$, that is the inverse modulus.

You can view the same congruences too, like to:
$$\begin{align*}
(p-5)! &\equiv (p-24)^{-1}&\pmod p\\
(p-4)! &\equiv (p+6)^{-1}&\pmod p\\
(p-3)! &\equiv (p-2)^{-1} &\pmod p\\
(p-2)! &\equiv 1&\pmod p\\
(p-1)! &\equiv -1&\pmod p\\
\end{align*}
$$
so, the sum is equal: $$ (-24)^{-1}+(6)^{-1} +(-2)^{-1}$$
and if you factorize in the beginning the factorials you get
$$ 8\cdot (-24)^{-1} \pmod p$$
And, voila, inverse modulus is more efficient than factorials.

StackExchange Computer Science Q#1495, answer score: 14