show numbers (mod $p$) are distict and nonzero Let's start with the nonzero numbers, mod $p$,
$1$, $2$, $\cdots$, $(p-1)$,
and multiply them all by a nonzero $a$ (mod $p$).
Notice that if we multiply again by the inverse of $a$ (mod $p$) we get back the numbers 
$1$, $2$, $\cdots$, $(p-1)$.
But my question is how the above process show that the numbers
$a\cdot 1$ mod $p$, $a\cdot 2$ mod $p$, $\cdots$, $a\cdot (p-1)$ mod $p$
are distinct and nonzero?
Thanks in advance. 

It is important to point out that $\Bbb Z_p$ is  a field if (and only if) $p$ is prime. This means in particular that every element that is not $0$ has an inverse. Since $p\not\mid a$ is equivalent to $a\not\equiv 0\mod p$, $a$ has an inverse $a^{-1}$. But then $$x\equiv y\mod p\iff ax\equiv ay\mod p$$
since we can reverse the equalities by multiplication by $a$ or $a^{-1}$. Moreover, since $p\not\mid a$ and $p\not\mid x$ (by assumption), $p\not\mid ax$, that is $$a\not\equiv 0,x\not\equiv 0\implies ax\not\equiv 0$$
