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(AB)'(AB) = I
(B'A')(AB) = [(B'A')A]B = [B'(A'A)]B = [B'I]B = B'B = I = (AB)'(AB)
So
(B'A')(AB) = (AB)'(AB)
Right multiply both sides by (AB)'
[(B'A')(AB)](AB)' = [(AB)'(AB)](AB)'
(B'A')[(AB)](AB)'] = (AB)'[(AB)(AB)']
(B'A')I = (AB)'I
B'A' = (AB)'
That's what we were to prove.
Edwin