Prove that $\sum_{j = 0}^p {p \choose j}{p + j \choose j} \equiv 2^p + 1 ~(\text{mod}~p^2)$
Let $p$ a prime number, then prove that : $\sum \limits_{k=0}^{p} \binom {p}{k} \binom{p+k}{k} \equiv 2^p +1 \pmod{p^2}$