**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.21**

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

Solution: 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 $(x^2)^k = x$.