TITLE: Proving that $3^{(3^4)}>4^{(4^3)}$ without a calculator
QUESTION [21 upvotes]: Is there a slick elementary way of proving that $3^{(3^4)}>4^{(4^3)}$ without using a calculator?
Here is what I was thinking:
$$4^4=256>243=3^5,$$
hence
$$4^{4^3}=4^{64}=(4^4)^{16}=(3^5)^{16}\cdot\left(\dfrac{256}{243}\right)^{16}=3^{80}\cdot\left(\dfrac{256}{243}\right)^{16}<3^{81}=3^{3^4}\,.$$
It is possible to prove that $\left(\dfrac{256}{243}\right)^{16}<3$ without a calculator by making a comparison such as $\dfrac{256}{243}=1+\dfrac{13}{243}<1+\dfrac{15}{240}=\dfrac{17}{16}$ and $\left(1+\dfrac{1}{16}\right)^{16}<e$ so $\left(\dfrac{256}{243}\right)^{16}<\left(1+\dfrac{1}{16}\right)^{16}<e<3$.
Is there a more elegant elementary proof? Ideally I'm looking for a proof that doesn't rely on calculus.
Source of problem: I made up this question but it is inspired by a similar question that appeared in the British Mathematical Olympiad round 1 in 2014.

REPLY [3 votes]: A bit of direct computation shows that
$$
\overset{\substack{531441\\\downarrow\\{}}}{3^{12}}\gt\overset{\substack{524288\\\downarrow\\{}}}{2^{19}}
$$
Squaring both sides gives
$$
3^{24}\gt4^{19}
$$
Since $3^{81/64}\gt3^{24/19}\gt4$, we get
$$
3^{81}\gt4^{64}
$$
which is the same as
$$
3^{3^4}\gt4^{4^3}
$$