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? 

I don't know Fraleigh's book, and I'm not near a library that has a copy right now so I can't check out his approach to elementary group theory, what he says, etc., but it looks like he wants to use "left cancellation", the property that $ab = ac$ implies $b = c$, to prove the only idempotent in a group is the identity element.  Left (and also right) cancellation of course hold in any group by virtue of the existence of inverses.  But is also possible to have cancellation in an algebraic structure without identity, for example the set $n \Bbb Z$ for integer $n > 1$, considering it equipped with only multiplication.  So perhaps Fraleigh is trying to show how this argument fits into a more general pattern.  Of course, if $G$ is finite and has an identity, then cancellation implies the existence of inverses, since in that case the map $g \to ag$ is injective, whence finiteness forces it to be surjective as well, so there must be some $b \in G$ with $ab = e$.  Then we could argue that $x^2 = x$ forces $x =e$, as shown by AWertheim in his answer, by simply multiplying by $x^{-1}$:  $x = x^{-1}x^2 = x^{-1}x = e$.  In any event, if one wants to proceed via cancellation, the equation  $x = xe$ is needed so that something is "left over" after one cancels out $x$!  It's not so much 
prognostication as it is experience with such 
maneuvers.  But it can 
seem a bit mysterious the first time you see it.
These things being said, it seems easier, clearer and cleaner to me to simply write
$x^2 = x \Rightarrow x^{-1}x^2 = x^{-1}x = e \Rightarrow x =e. \tag{1}$
Hope this helps!  Merry Christmas to One and All,
and as always,
Fiat Lux!!!
