Fermat’s theorem proof

Consider the sequence $a,\mathrm{\hspace{0.17em}2}a,\mathrm{\dots },(p-1)a$.

They are all different (modulo $p$), because if $ma=na$ with then$0=a(m-n)$, and since  $p\nmid a$, we get $p\mid (m-n)$,  which is impossible.

Now, since all these numbers are different, the set  $\{a,\mathrm{\hspace{0.17em}2}a,\mathrm{\hspace{0.17em}3}a,\mathrm{\dots },(p-1)a\}$  will have the $p-1$ possible congruence classes (although not necessarily in the same order) and therefore

$$a\cdot 2a\cdot 3a\mathrm{\cdots }(p-1)a\equiv (p-1)!a^{p-1}\equiv (p-1)!(modp)$$

and using  $\gcd ((p-1)!,p)=1$  we get

$$a^{p-1}\equiv 1(modp).$$