polynomials - Number Theory: Prove that $x^{p-2}+dots+x^2+x+1equiv 0pmod{p}$ has exactly $p-2$ solutions

I just completed this homework problem, but I was wondering if my proof was correct:

If $p$ is an odd prime, then prove that the congruence $x^{p-2}+\dots+x^2+x+1\equiv 0\pmod{p}$ has exactly $p-2$ incongruent solutions, and they are the integers $2,3,\dots,p-1$.

Proof

Let $f(x)=x^{p-2}+\dots+x^2+x+1$ and $g(x)=(x-1)f(x)$.

Note that $f(1)=(p-2)+1=p-1\not\equiv 0\pmod{p}$.

So, $g(x)=(x-1)(x^{p-2}+\dots+x^2+x+1)=x^{p-1}-1$.

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 $2,3,\dots,p-1$. $\blacksquare$