Is there a simple proof that the product of a simple k-vector with its reverse always yields a scalar? Assume real Clifford algebra of arbitrary but finite dimension with unspecified metric. By simple k-vectors I mean elements of the algebra that can be written as the outer product of k vectors, e.g. $a\wedge b\wedge c$, whose reverse is $c\wedge b\wedge a$.
The wikipedia page http://en.wikipedia.org/wiki/Clifford_algebra#Relation_to_the_exterior_algebra talks about natural isomorphism between Clifford and exterior algebra. Accordingly, some elements of Clifford algebra are isomorphic to elements of exterior algebra that have the form $a\wedge b$, $a\wedge b\wedge c$, etc. These elements are blades in exterior algebra, so I call them blades in Clifford algebra as well.
Feel free to use whatever definition of Clifford algebra is most convenient for the proof.