why does this approach neglect the output resistance of the bjt in calculating gain?
Surely, if Vc is equal to Vcc- Rc(ic) then that is regarding the
  circuit as DC and ignoring the effect of the output resistance of the
  bjt


It's simple ohms law; if a current \$I_C\$ flows through \$R_C\$ then the volt drop it creates subtracts from \$V_{CC}\$ to give you \$V_C\$. You can only regard the output resistance of the collector being effectively in parallel with \$R_C\$ when doing an AC small signal analysis.