\(1^{-1}\cdot 2^{-1} + 2^{-1} \cdot 3^{-1} +... + (p-2)^{-1}\cdot (p-1)^{-1}\\ =\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{(p-1)(p-2)}\\ =(1-\dfrac{1}{2})+(\dfrac{1}{2}-\dfrac{1}{3})+...+(\dfrac{1}{p-1}-\dfrac{1}{p-2})\\ =1-\dfrac{1}{p-2}\\ =\dfrac{p-3}{p-2}\)

So our work is to evaluate

\(\dfrac{p-3}{p-2}\pmod p\).

Don't know how to do that, if I were you, I would try some primes.

So let's try it.

When p = 7,

\(\dfrac{p-3}{p-2}\pmod p\\ =\dfrac{4}{5}\mod 7\\ =\dfrac{12}{15}\mod 7\\ =5\)

When p = 11,

\(\dfrac{p-3}{p-2}\pmod p\\ =\dfrac{8}{9}\pmod {11}\\ =\dfrac{40}{45}\pmod {11}\\ =7\)

When p = 13,

\(\dfrac{p-3}{p-2}\pmod p\\ =\dfrac{10}{11}\pmod {13}\\ =\dfrac{60}{66}\pmod {13}\\ =8\)

When p = 17,

\(\dfrac{14}{15}\pmod{17}\\ =\dfrac{112}{120}\pmod {17}\\ =10\)

Welp can't see a pattern :(