Proving that $3^{(3^4)}>4^{(4^3)}$ without a calculator 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.

$3^4-4^3=17$ therefore we can rewrite the inequality as:
$$3^{17} > \left(\frac{4}{3}\right)^{4^3}$$
we have $\left(\frac{4}{3}\right)^8 < 10$, hence it suffices to show:
$$3^{17} > 10^8$$
or
$$3 \cdot 81^4 > 10^8$$
which is satisfied if:
$$3 \cdot 80^4 > 10^8$$
or
$$3 \cdot 2^{12} \cdot 10^4 > 2^{4} \cdot 5^4 \cdot 10^4$$
$$3 \cdot 2^8 > 5^4$$
i.e.
$$768 > 625$$
ADDENDUM
$4^8=4^4 \cdot 4^4 = 256 \cdot 256$ and $3^8=3^4 \cdot 3^4 = 81 \cdot 81$, it is not too difficult to make the two multiplications and then the division.
Also, if you are a programmer, you know already that $4^8 = 2^{16} = 65536$, so you just need to make $81 \times 81$ and then the division. Rather, you don't need any division because you see immediately $81 \cdot 81 = 6561 \gt 65536 / 10$.
Anyway, I admit, I have used the calculator :-)
