A number theory problem by Amiya Panda

We first note that $97$ is a prime and as such, by Wilson's Theorem, we have,

$96!\equiv (-1)\pmod{97}$

Now, $95!=1\times 2\times 3\times \cdots 94\times 95$ . Applying modulo $97$ , this can be rewritten as,

$95!\equiv (-96)\times (-95)\times \cdots \times (-2)\equiv -(96!)\pmod{97}$

Using the first result, we can write it as $95!\equiv -(-1)\equiv 1\pmod{97}$

$\therefore 95!\equiv 1\pmod{97}$

(p-2)!/p=1