Number Theory dealing with modulo 
I am having a hard time understanding the last part stating that states "the result follows from comparing coefficients". How can you conclude that the coefficients must be 0 (mod p)?
 A: First of all, we are subtracting the $2$ polynomials modulo $p$.
$$P_1(x)=x^{p-1}-S_1x^{p-2}+S_3x^{p-3}- \ldots-S_{p-2}x+S_{p-1} = 0 \pmod p$$
Has $p-1$ roots and so does the polyonomial :
$$P_2(x)=x^{p-1}-1 = 0\pmod p$$
Subtracting these both, 
$$P_1(x)-P_2(x)=-S_1x^{p-2}+S_3x^{p-3}- \ldots-S_{p-2}x+S_{p-1}+1 = 0 \pmod p$$
we get a polynomial having $p-1$ roots, but it's a polynomial with degree $p-2$.
This implies that the polynomial is none else than the polynomial $P(x)=0 \pmod p$
Now, since this has become an identity that $$\forall x \in \mathbb R~~ -S_1x^{p-2}+S_3x^{p-3}- \ldots-S_{p-2}x+S_{p-1}+1 = 0 \pmod p$$
This implies that $p \mid S_i ~\forall i \in \{1,2, \ldots , p-2 \}$
Also, $S_{p-1}+1=0 \pmod p$ because $S_{p-1}$ is none else than $(p-1)!$, and by Wilson's Theorem : $$(p-1)!+1 = 0\pmod p$$
A: What it's trying to say is that:
$$(x-1)(x-2)...(x-(p-1))=x^{p-1}-1$$
in $Z_p[X]$ - that's because they have the same roots, their degrees are less than $p$ and have the same leading coefficient. Because they are equal as polynomials, their coefficients are also equal and thus you get $S_i \equiv 0$ for $i\le p-1$.
