User blog comment:D57799/Ranging FGH before omega/@comment-5529393-20141008144817

Nice proof! Here's a proof that 2 ^{m-1} 2n >= 2f_m (n) for m,n >= 1:

Here ^{m} indicates ^...^ with m arrows, and (2 ^{m})^n indicates 2 ^ {m} 2 ^{m} ... 2 ^{m} with n copies of 2 ^ {m} (and similarly for (f_m)^n)

Lemma: 2^{m} (n) >= 2n for m >= 0, n >= 2

Proof of Lemma:

2^{m} (2-1) = 2

2^{0} n = 2n

2^{m} (n+1) = 2 ^ {m-1} (2 ^{m} n) >= 2^{m-1} 2n = 4n > 2(n+1)

Main Proof:

m = 1: 2 * 2n = 4n = 2 f_1(n)

n = 1: 2 ^{m-1} 2(1) = 4 = 2 f_m (1)

Main Induction step:

2 f_m(n) = 2 (f_{m-1})^n (n) <= (2^{m-2})^n (2n) <= (2^{m-2})^n (2^{m-1} n) = 2^{m-1} (n). QED