Modulus proof with factorials

Question

Prove that $(2p−1)! ≡ p \pmod {\space p^2}$
Hint: $p$ divides ($2p-1)!$

Any help would be much appreciated.

Answer 1

The case $p=2$ is easy to check by hand, hence we may assume $p>2$. $$\frac{(2p-1)!}{p}= 1\cdot 2\cdots (p-1)\cdot(p+1)\cdot(p+2)\cdots (2p-1)=\prod_{k=1}^{p-1}(p^2-k^2) $$ is equivalent $\!\!\pmod{p}$ to: $$ (-1)^{p-1}\left[(p-1)!\right]^2 $$ that by Wilson's theorem is $1\pmod{p}$.

Answer 2

Just to give another way. Noticed that $$(p+1)(p+2)....(2p-1)=mp^2+(p-1)!$$ It follows, using Wilson, $$(2p-1)!=(p-1)!p(p+1)(p+2)....(2p-1)=p(pn-1)(mp^2+(p-1)!)=(p^2n-p)(mp^2+pn-1)=Mp^2+p$$

Thus $$(2p−1)! ≡ p \pmod {\space p^2}$$