Prove Wilson's theorem by Lagrange's theorem

[kingwinner]

Lagrange's Theorem: let p be any prime and f(x) = a_nx^n +a_{n-1}x^{n-1} + ... + a_1x + a_0 with a_n ≡/≡ 0 (mod p). Then f(x) ≡ 0 (mod p) has at most n solutions.

Use the above theorem to prove Wilson's theorem.
Hint: Let f(x) = (x-1)(x-2)...(x-(p-1)) - (x^{p-1} - 1) for an odd prime p.

Proof:
Expanding,
f(x) = (x-1)(x-2)...(x-(p-1)) - (x^{p-1} - 1)
= a_{p-2}x^{p-2} +...+ a_1x + a_0 where the a_i are some coefficients
By the above theorem, f has at most p-2 roots mod p IF a_{p-2} ≡/≡ 0 (mod p). (*)
But by Fermat's theorem, for a=1,2,...,p-1, a^{p-1} -1 ≡ 0 (mod p).
So for a=1,2,...,p-1, f(a) ≡ 0 (mod p).
So f has at least p-1 roots mod p. (**)
(*) and (**) contradict unless f(x) ≡ 0 (mod p). Therefore, we must have f(x) ≡ 0 (mod p).
=> f(0)=(-1)(-2)...(-(p-1)) - (-1) ≡ 0 (mod p)
=> (-1)^{p-1} (p-1)! + 1 ≡ 0 (mod p)
For odd primes p, p-1 is even, so (p-1)! ≡ -1 (mod p)
(For p=2, check directly.)
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I don't understand the two lines in red.
I understand that there is a contradiction, but why does this imply that f(x) ≡ 0 (mod p)? Why in this case, there will be no contradiction? I'm totally lost here...

Also, f(x) ≡ 0 (mod p) doesn't necessarily mean it holds for EVERY integer x, so why can we substitute x=0 and say that f(0) ≡ 0 (mod p)? What is the justification for this step?

I hope someone can explain this proof.
Thank you very much!

[also under discussion in math help forum]

 

[JSuarez]

If a polynomial with at most p-2 roots has, in fact, p-1, then it must be identically 0.

 

[kingwinner]

If a polynomial with at most p-2 roots has, in fact, p-1, then it must be identically 0.

Why? What is the reasoning?
Polynomial with at most p-2 roots has, in fact, p-1 roots <----this is already a contradiction. The statement is inconsistent...

 

[JSuarez]

Because the complete statement of Lagrange's theorem, is that, if f(x) is a nontrivial (not identically 0) polynomial with degree n, then it has at most n roots (mod p).

This doesn't happen with your polynomial, so it must be the trivial one. What does this imply for f(0)(mod p)?

 

[kingwinner]

Thanks! I think I got it!