number theory - How to show $x^{p-2}+cdots+x^2+x+1equiv0pmod{p}$ has $p-2$ incongruent solutions: $2, 3,..., p-1.$ when $p$ is an odd prime?

Let $p$ be an odd prime. How would I prove that the congruence $x^{p-2}+\cdots+x^2+x+1\equiv0\pmod{p}$ has exactly $p-2$ incongruent solutions, and they are the integers $2, 3,..., p-1?$

I found this question posted, but the answer doesn't quite make sense to me.

Here is my attempt:

Proof
As $p$ is an odd prime, by Fermat's Theorem,
\begin{align*}
x^{p-1}&\equiv1\pmod{p}\\
&\implies x^{p-1}-1\equiv0\pmod{p}
\end{align*}
and by Lagrange's Corollary, with $d=p-1, p-1\mid p-1$, this congruence has exactly $p-1$ solutions.
\begin{align*}
x^{p-1}-1 \pmod{p}&\equiv0\pmod{p}\\

&\equiv (x-1)(x^{p-2}+\cdots+x^2+x+1) \pmod{p}\\
\end{align*}
Assume $x^{p-2}+\cdots+x^2+x+1\equiv0\pmod{p}$...

And then I get stuck.

The proof I linked above used:

Now, $x^{p-1}-1\equiv 0\pmod{p}$ has exactly $p-1$ incongruent solutions modulo $p$ by Lagrange's Theorem.

Note that $g(1)=(1-1)f(1)=0\equiv 0\pmod{p}$, so $1$ is a root of $g(x)$ modulo $p$. Hence, the incongruent roots of $g(x)$ modulo $p$ are $1,2,3,\dots,p-1$.

But every root of $g(x)$ other than $1$ is also a root of $f(x)$ (This is the part I'm concerned about. Is it clear that this is the case?), hence $f(x)$ has exactly $p-2$ incongruent roots modulo $p$, which are $2,3,\dots,p-1$. $\blacksquare$

But I don't understand the part about roots. Nor do I quite get the use of creating $g(x)$ and $f(x)$. Could someone explain the proof using a different method?