Powers Up to Order In Group Are Distinct

Proof

Let $g\in G$ be an element of order $n$. Then, consider $g^i$ and $g^j$ for some $i,j\in \{0,1,2,\dots ,n-1\}$, such that $g^i=g^j$.

Then, we have that $g^{i-j}=\mathrm{id}$.

However this implies that $i-j$ must be a multiple of $n$, but $i,j\in \{0,1,2,\dots ,n-1\}$, which leaves only the possibility that $i-j=0$.

Therefore we have that $g^i=g^j⟹i=j$ and therefore by the contrapositive $i\ne j⟹g^i\ne g^j$.

In the infinite case, we know that if $g^i=g^j$ then similarly $i-j=0$, since no non-zero power of $g$ is the identity.