Let $p$ be an odd prime. Prove that $$\sum_{j = 0}^p {p \choose j}{p + j \choose j} \equiv 2^p + 1 ~(\text{mod}~p^2)$$
I have tried to write the problem as $$\sum_{j = 0}^p {p \choose j}{p + j \choose j} - 2^p - 1 \equiv 0~(\text{mod}~p^2)$$ Which mean I have to prove that the expression on the left side is divisible by $p^2$, but this doesn't help.
I also know that since $p$ is odd prime, we have $2^{\varphi(p^2)} = 2^{p^2 - p} \equiv 1 ~(\text{mod} ~p^2)$ by Euler's theorem. So the expression before can be rewritten as $$\sum_{j = 0}^p {p \choose j}{p + j \choose j} - 2^p - 2^{p^2 - p} \equiv 0~(\text{mod}~p^2)$$ But this also doesn't seem to help. I think I have to modify the sum somehow, but I gut stuck. The sum looks like a hypergeometric function, but knowing this doesn't help either.
Anyone?