**Proof.** Let $G$ be a group, and $x,y\in G$. Then, we have $xy(xy)^{-1}=e$, where $e$ is the identity element of $G$. Multiplying on the left by $x^{-1}$, we get $y(xy)^{-1}=x^{-1}$. Multiplying once again on the left by $y^{-1}$ gives $(xy)^{-1}=y^{-1}x^{-1}$, which was what was to be shown.

Obviously this comes to the correct conclusion, but is it valid by the laws of algebra? The assumption that $xy(xy)^{-1}=e$ is directly from the fact that $G$ is a group (closed under inversion and the operation), but perhaps it's just my tendency to view $xy$ as two completely separate elements rather than an actual member of $G$ that's distracting me?

Is there a better way to show this? Because if it is valid, it seems (to me, at least, and beauty of the proof is in the eye of the writer..) more "intuitive" than the proof my book gave.