TITLE: Prime numbers with binomial coefficients
QUESTION [5 upvotes]: Question:
Prove that for any prime $p>3$, the number $\binom{2p-1}{p-1}-1$ is divisible by $p^{3}$.
Attempt:
Since every integer that is relatively prime to p has a multiplicative inverse
modulo p, denote the inverse of x modulo p by x−1. We start by improving the
conclusion of the previous problem. Therefore:
$$\binom{2p}{p}-2=\sum_{k=1}^{p-1}\binom{p}{k}^{2}=\sum_{k=1}^{p-1}\left ( \frac{p}{k}\binom{p-1}{k-1} \right )^{2}$$
I feel like I have made a mistake because unless $\frac{1}{k}\binom{p-1}{k-1}$ is an integer I can't proceed any further. Is it an integer? Where should I go from here?

REPLY [3 votes]: The expression eventually turns out to be $\dfrac{1}{2}$ $ 2p \choose p$$-1$ from here .
Wolstenholme's theorem: 
$ \dfrac{1}{2}{2p\choose p}\equiv 1\mod p^3$
Just Collecting the facts.