Let $p$ be prime. Prove that $p$ divides $(1− n) (1 + n + n^2 + n^3 + ⋯ + n^{p−2})$, provided $p$ does not divide $n$.
We first expand and simplify the expression
$$ \begin{align} (1− n) (1 + n + n^2 + n^3 + ⋯ + n^{p−2}) & = 1 + n + n^2 + n^3 + ⋯ + n^{p−2} \\ & - n - n^2 - n^3 - \ldots - n^{p-1} \\ \\ & = 1 - n^{p-1} \end{align} $$
Since $p$ does not divide $n$, we can use the FlT,
$$ n^{p-1} \equiv 1 \pmod p $$
And so
$$ \begin{align} (1− n) (1 + n + n^2 + n^3 + ⋯ + n^{p−2}) & \equiv 1 - n^{p-1} \pmod p \\ \\ & \equiv 1 - (1) \pmod p \\ \\ & \equiv 0 \pmod p \end{align} $$
This tells us that $p$ divides $(1− n) (1 + n + n^2 + n^3 + ⋯ + n^{p−2})$, if $p \not \mid n$.