For any prime $p>3$ show that $C_{np}^{p}-C_{np}^{2p}+C_{np}^{3p}-C_{np}^{4p}+...+(-1)^{n-1}C_{np}^{np} \equiv 1\pmod{p^3}$

Question

Let $n$ be a positive integer. For any prime $p>3$ show that $$C_{np}^{p}-C_{np}^{2p}+C_{np}^{3p}-C_{np}^{4p}+...+(-1)^{n-1}C_{np}^{np} \equiv 1\pmod {p^3}$$ Where $C_{n}^{k}=\frac{n!}{k!(n-k)!}$. (*)

I tried this by brute force - plugging (*) everywhere and simplify, but I couldn't make it produce the desired result. It would be appreciated if somebody would show me some tricks for a better approach.

Answer 1

According to Binomial Congruence (see also Bailey's paper) we have that for any prime $p\geq 5$, $$\dbinom{np}{kp}\equiv\binom{n}{k}\pmod {p^3}.$$ Hence $$\sum_{k=1}^n (-1)^{k-1}\binom{np}{kp}\equiv-\sum_{k=1}^n (-1)^k\binom{n}{k}=-(1-1)^n+\binom{n}{0}=1\pmod {p^3}.$$