While looking for alternative proofs for the theorem, I came across the following link
The proof
Let $p$ be a prime number. By Fermat's little theorem, all non-zero elements of the field must be the roots of the polynomial $P(x)=x^{p-1}-1$.
$x^{p-1}-1= \prod_{r = 1}^{p-1}(x-r)$
Now, either $p=2$, in which case $a \equiv -a \pmod 2$ for any integer $a$, or $p-1$ is even. In either case, $(-1)^{p-1} \equiv 1 \pmod{p}$, so that
$x^{p-1}-1=\prod_{ r=1}^{p-1}(x-r)=\prod(r-x)$
If we set $x=0$ then we get the theorem.
My question is, if we have assumed that $x$ is non-zero in the beginning, how can we substitute it back in the end to get the theorem?
Can someone verify if this proof is correct or not and, if it is, explain why doing this is allowed?
It is a special case of the following
Theorem $ $ Suppose $f(x)$ and $g(x)$ are polynomials of degree $n$ whose coefficients lie in a field $F$ (e.g. $\Bbb Q,\, \Bbb R$ or $\Bbb Z_p =$ integers $\bmod p$). If $f$ and $g$ have equal degree $n$ and equal lead coefs and they have equal values at $n$ distinct points $\in F$ then they have equal corresponding coefs $\,f_i = g_i\,$ for all $\,i\,$ (i.e. $f = g$ as formal polynomials), therefore $f$ and $g$ have equal values at all points $\in F$.
Proof $\ $ By hypothesis $f$ and $g$ have equal lead terms which cancel in $\,h := f-g\,$ so $\,\color{#c00}{\deg h < n}.\, $ Thus $\,h\,$ is zero at the $\color{#c00}{n > \deg h}\,$ distinct points where $f$ and $g$ have equal values. By a well-known theorem, if a polynomial over a field has $\rm\color{#c00}{more\ roots}$ than its degree then it is the zero polynomial, i.e. all its coefs are zero, so $\, h_i = f_i - g_i = 0$.