Assume we have $\mathbb{Z}_{p}[x]$ with $p$ being a prime. Prove that $x^{p-1}-1=(x-1)(x-2)...(x-(p-1))$

Question

I know how this formula works and it is quite interesting actually but how would you prove this relationship? Through induction (seems difficult since there's no equation for prime numbers), but I'm not sure if there is any other way to show this equation. Any ideas?

Answer 1

Hint: $\mathbb{Z}_p^* = \mathbb{Z}_p \setminus \lbrace 0 \rbrace$ is a group of order $p-1$.

Answer 2

Recall the simple fact that if $gcd(x,a)=1$ then, $x^{\phi(a)} \equiv 1 \ (\text{mod a})$.

Here, $gcd(x,p)=1 \Rightarrow x^{\phi(p)} \equiv 1 \ (\text{mod p}) \Rightarrow x^{p-1} \equiv 1 \ (\text{mod p}) \Rightarrow \forall z \in \mathbb{Z}_p,\ z^{p-1}-1 =0$. This is enough to factor the polynomial in to a product of linear factors where all of it's roots lie in $\mathbb{Z}_p$.