Question:
Given that $p\ge 7$ is a prime number, evaluate $$1^{-1} \cdot 2^{-1} + 2^{-1} \cdot 3^{-1} + 3^{-1} \cdot 4^{-1} + \cdots + (p-2)^{-1} \cdot (p-1)^{-1} \pmod{p}.$$

Answer:
As $p$ is a prime number, it follows that the modular inverses of $1,2, \ldots, p-1$ all exist. We claim that $n^{-1} \cdot (n+1)^{-1} \equiv n^{-1} - (n+1)^{-1} \pmod{p}$ for $n \in \{1,2, \ldots, p-2\}$, 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 $$1 \equiv n(n+1) \cdot (n^{-1} - (n+1)^{-1}) \equiv (n+1) - n \equiv 1 \pmod{p},$$as desired. Thus, \begin{align*}&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} \pmod{p}.\end{align*}This is a telescoping series, which sums to $1^{-1} - (p-1)^{-1} \equiv 1 - (-1)^{-1} \equiv \boxed{2} \pmod{p}$, since the modular inverse of $-1$ is itself.