Fermat’s little theorem states that given any positive prime $p$ and any integer $x$, $x^p-x$ is always divisble by $p$. First we take a look at the special cases:
When $p=2$, $x^2-x=x(x-1)$ is always divisble by $2!$.
When $p=3$, $x^3-x=x(x^2-1)=(x-1)x(x+1)$ is always divisble by $3!$.
When $p=5$, $x^5-x=(x-1)x(x+1)(x^2+1)$ and is congruent to $(x-1)x(x+1)(x^2-4)=(x-2)(x-1)x(x+1)(x+2)$ modular $5$ is divisble by $5!$.
When $p=7$, $x^7-x=x(x-1)(x^2+x+1)(x^3+1)$ is congruent to $x(x-1)(x^2+x-6)(x^3+8)=x(x-1)(x+3)(x-2)(x+2)(x^2+2x-4)$ is congruent to $x(x-1)(x+3)(x-2)(x+2)(x^2+5x+4)=x(x-1)(x+3)(x-2)(x+2)(x+1)(x+4)$ modular $7$ thus is divisble by $7!$.
Do we have the same pattern for all $x^p-x$ when $p$ is prime and can we give another proof to Fermat’s little theorem using this way?