Suppose there's an element in G of order 2n-1 that is it's own inverse and try to prove it's a contradiction.
by a corollary to lagrange, the order of an element divides the order of G. if an element is it's own inverse, it is either the identity, or of order 2.
as you have stated your problem, the statement is false, since EVERY group contains an identity element, which is ALWAYS it's own inverse.
what IS true, is that every non-identity element of a group of order 2n-1 cannot be its own inverse.
because 2n-1 is odd, and 2 does not divide an odd number.