jmc

As $p$ is a prime number, it follows that the modular inverses of $1,2,\dots ,p-1$ all exist. We claim that $n^{-1}\cdot (n+1{)}^{-1}\equiv n^{-1}-(n+1{)}^{-1}(\mathrm{mod}p)$ for $n\in \{1,2,\dots ,p-2\}$, in analogue with the formula $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$. Indeed, multiplying both sides of the congruence by $n(n+1)$, we find that $$1\equiv n(n+1)\cdot (n^{-1}-(n+1{)}^{-1})\equiv (n+1)-n\equiv 1(\mathrm{mod}p),$$as desired. Thus, $$\begin{array}{rl}& 1^{-1}\cdot 2^{-1}+2^{-1}\cdot 3^{-1}+3^{-1}\cdot 4^{-1}+\cdots +(p-2{)}^{-1}\cdot (p-1{)}^{-1}\\ & \equiv 1^{-1}-2^{-1}+2^{-1}-3^{-1}+\cdots -(p-1{)}^{-1}(\mathrm{mod}p).\end{array}$$This is a telescoping series, which sums to $1^{-1}-(p-1{)}^{-1}\equiv 1-(-1{)}^{-1}\equiv \boxed{2}(\mathrm{mod}p)$, since the modular inverse of $-1$ is itself.