Euler's theorem: https://www.cnblogs.com/wangxiaodai/p/9758242.html
Factorial inverse:
记 f[i] = i! mod p,
g[i] = (i!)−1 mod p
Easy to find g [i] = g [i ] + 1 * (i +1)
i−1 = f[i]−1∗g[i]
Only calculating f [n], is obtained and f [n] The inverse element g [n], and can be recursive.