TITLE: Prove that $6<3^\sqrt3$ implies $3^\sqrt3 < 7$
QUESTION [16 upvotes]: Consider the double inequality : $$6<3^\sqrt3 < 7$$
  Using $only$ the elementary properties of exponents and inequalities (NO calculator, computer,
  table of logarithms, or estimate of √3 may be used), prove that the first inequality implies the
  second.
SOURCE : "Inequalities proposed in Crux Mathematicorum" (Page Number 14; Question Number 627)

I have no idea how to solve this problem. I tried taking logarithms, but that did not help. I strongly suspect that some inequality has to be used. I tried AM-GM on a few set of terms, but that just made it messier.
Wolfram Aplha gives the value of ${3^{\sqrt {3}}}$ as $\approx 6.7049918538258$.
Can anyone give me a hint on how to solve this problem ?

REPLY [16 votes]: Claim: $7 > 3^{\sqrt 3}$. Otherwise
$$7^\sqrt 3 \le 27 < 28 = 7*4 \\
7^{\sqrt 3 - 1} < 4 \\
48 < 7^2 < 4^{\sqrt 3 + 1} \text{(since }(7^{\sqrt 3 - 1})^{\sqrt 3 + 1} = 7^{(\sqrt 3 - 1)(\sqrt 3 + 1)} = 7^{(\sqrt 3)^2 - 1^2} = 7^2{)} \\
3 < 4^{\sqrt 3 - 1} \: \text{(divide by 16)}\\
3^{\sqrt 3 + 1} < 4^2 = 16 < 18 \\
3^\sqrt 3 < 6$$
...and we have a contradiction with a given statement.