Trick to proving a group has exactly one idempotent element - Fraleigh p. 48 4.31 

If $*$ is a binary operation on a set $S$, an element $x \in S$ is an idempotent for $*$ if $x * x = x$.

Let $\langle G, *\rangle$ be a group and let $x\in G$ such that $x*x = x.$ Then $x*x = x*e$, and by left cancellation,  $x = e$, so $e$ is the only idempotent element in a group.

The trick here looks like writing $x$ as $x*e$. How can you prognosticate (please see profile) this? I didn't see it. It also looks like you have to prognosticate the 'one idempotent element' to be the identity element. Is this right? Can someone make this less magical and psychic? 

There are two things I want to address here:
1) There aren't that many things we can actually do. We know group elements are closed under an operation, the operation is associative, and that every element has an inverse element, and that there is an identity for the group. What could the single, unique idempotent element be? It has to be the identity.
2) We multiply things by identities all the time. Think about rationalizing denominators, or finding GCDs to add fractions. You'll probably even do it again in regards to groups, since $gg^{-1} = 1$ could be helpful at some point.
