$g$ has order $n$, then $\langle g\rangle=\langle g^2\rangle=\cdots=\langle g^{n-1}\rangle$

Question

If $g$ has order $n$, then $\langle g\rangle=\langle g^2\rangle=\cdots=\langle g^{n-1}\rangle$.

This should be fairly easy but somehow I just couldn't prove it. I only managed to prove the case when $n$ is prime (which might be wrong, I haven't checked carefully). Say $x=g^k$ for $k<n$ and we want to show that it is in $\langle g^m\rangle$, then we want to write $x$ as $g^{mp}$, but then I don't see how to proceed. Any hint?

Update: so I found a counterexample when $n$ is not prime. Sorry for the confusion. Anyway, if anyone could provide a proof when $n$ is prime, still appreciated.

Answer 1

Hint. If $ n $ is prime, given any $ k < n $, we have ${\rm gcd} (k, n)= 1$. Hence, we can write $1= an + bk $ for some $ a, b \in \mathbf Z $. Then $$ g = (g^n)^a \cdot (g^k)^b. $$