To prove ${2p - 1 \choose p } \equiv 1 \pmod{p^2}$ without using Wolstenholme's theorem

Question

How to prove that ${2p - 1 \choose p} \equiv 1 \pmod{p^2}$ ? I don't want to use Wolstenholme's theorem; but one might use $p|{p \choose k} , 1 \le k \le p - 1$ , and $(p - 1)! \sum_{k = 1}^{p - 1} \dfrac 1k \equiv 0 \pmod p$. Please help.

Answer 1

We have: $$(p-1)!\cdot\binom{2p-1}{p}=\prod_{k=1}^{p-1}(p+k)= (p-1)!\cdot(1+p H_{p-1})+Kp^2\tag{1}$$ for some $K\in\mathbb{N}$. Now it is well known that $H_{p-1}\equiv 0\pmod{p}$.

Since $(p-1)!\equiv -1\pmod{p}$, we have that $(p-1)!$ is invertible $\pmod{p^2}$,

hence the previous line easily gives: $$\binom{2p-1}{p}\equiv 1\pmod{p^2}\tag{2}$$ as wanted.