This last thing is not hard to prove 
2^{k+n} + 2 \times 3^{2k+1} - 7 \times 6^k > 2^k + 2 \times 3^{2k+1} - 7 \times 6^k = 2^k + 6 \times 3^{2k} - 7 \times 6^k
Hence we want: 2^k + 6 \times 3^{2k} > 7 \times 6^k
Which in logs, rewrites as: 5 \times \text{log}_2(3) > 6 \times \text{log}_2(2)
Which is true: 5 \times \text{log}_2(3) \simeq 7.925
Hence, if everything is correct above this message we have an elementary proof of the inequality, of the < 2^n bound and of Elementary proof of no circuits?!!