Prove that $(p-1)(p-2)\cdots(p-r)\equiv(-1)^rr!(mod\, p) $ for $r=1,2\cdots,(p-1)$

Question

Prove that $(p-1)(p-2)\cdots(p-r)\equiv(-1)^rr!(mod\, p) $ for $r=1,2\cdots,(p-1)$

MY TRY:I know Wilson's lemma but here ,I am clueless to solve this problem.Thank you

Answer 1

Hint:

$$(p-1)\equiv-1\pmod p$$

$$(p-2)\equiv-2\pmod p$$

$$(p-3)\equiv-3\pmod p$$

$$\text{etc.}$$

Multiply the result together and see what you get.

Answer 2

Hint: Expand the product and note that any summand containing a factor $p$ gives a contribution of $0$.

Of course you can use induction to formalise the argument.