Prove that if $[a]^s = [1]$ in $\mathbb{Z}_m$, for some $s \in \mathbb{N}$, then $[a]^t \not = [0]$, for all $t \in \mathbb{N}$

Question

Prove that if $[a]^s = [1]$ in $\mathbb{Z}m$, for some $s \in \mathbb{N}$, then $[a]^t \not = [0]$, for all $t \in \mathbb{N}$.

So far my approach are as follows

When trying the forward direction all i get is that $a^s-1=m(x), x \in \mathbb{Z}$. Then you can write $1$ as a linear combination of $a$ and $m$ so thus $a$ and $m$ are relatively prime. Thus $[a] \not = 0$ so $[a]^t \not =0$

With the contrapositve I tried

So if $[a]^t = [0]$ implies that $[a]=0$ thus $[a]^s = [1]$ in $\mathbb{Z}m$ is only true if $m=1$. I didn't properly deal with the quantifiers though so I don't think this is correct.

Answer 1

It is not because $[a]\ne 0$ that $[a]^t\ne 0$. For instance, in $\mathbf Z/8\mathbf Z$, $[6]^3=0$, albeit $[6]\ne0$.

You can observe that if $[a]^s=1$, $[a]$ is a unit in the ring $\mathbf Z/m\mathbf Z$, with inverse $[a]^{s-1}$. Suppose $[a]^{t}=0$ for some $t$.

You have $[a]^{t}=[a]^{t\bmod s}$, so you can suppose $t<s$. But then $$[a]^{s}=1=[a]^{t}[a]^{s-t}=0\cdot[a]^{s-t}=0,$$ which is a contradiction.