Every group element of odd order is an odd power of its square

Let $G$ be a finite group and let $g \in G$ be an element of order $n$. Prove that if $n$ is odd, then $x = (x^2)^k$ for some $k$.

Since $n$ is odd we have $n = 2k - 1$ for some integer $k$. Then $x^n = x^{2k-1} = x^{2k} x^{-1} = 1$, hence $(x^2)^k = x$.