Hard Inequality Problem: Prove that $6<3^{\sqrt{3}}<7$

Prove that $6<3^{\sqrt{3}}<7$ from the simple and straightforward inequality where $1<\sqrt{3}<2$.

This is a particularly daunting mathematics challenge and you could spend days or even a week trying to prove the inequality, with no fruitful result.

Notice that if we exponentiate the given inequality $1<\sqrt{3}<2$ with base 3, we get:

$3^1<3^{\sqrt{3}}<3^2$

$3<3^{\sqrt{3}}<9$ (Compare it with the targeted inequality $6<3^{\sqrt{3}}<7$)

The lower and upper bound that we could get from the given inequality is far too low and high for the targeted inequality.