G is a finite group and $\displaystyle x\in G$ and $\displaystyle |x|=m$. Prove that if m is odd, then $\displaystyle x=(x^2)^k$ for some $\displaystyle k\in\mathbb{Z}$, $\displaystyle k\geq 1$

$\displaystyle x^m=x^{2k+1}=(x^{2})^kx$

I need k to be 0 though. What am I missing?