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.
 A: Your problem is equivalent to proving that $\log_2{3} > \frac{128}{81}$.
One approach is to calculate that $3^{12} = 81^3 = 531441 > 2^{19} = 2^{10} \times 2^9 = 1024 \times 512 = 524288$.  Admittedly, this is a bit tedious with pencil-and-paper arithmetic, but it's doable.  From this, we get $\log_2{3} > \frac{19}{12}$.
Separately, calculate that $\frac{19}{12} = \frac{513}{324} > \frac{128}{81} = \frac{512}{324}$.
Combining these results, using the fact that the > operator is transitive, we get $\log_2{3} > \frac{128}{81}$, Q.E.D.
A: $3^4-4^3=17$ 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 :-)
A: 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
$$
3^{3^4}\gt4^{4^3}
$$
A: $$3^{81}\gt 2^{128}\iff 3\times 9^{40}\gt 256\times 8^{40}\iff (9/8)^{40}\gt \frac{256}3$$
Now,
$$(9/8)^{40}=(1+1/8)^{40}$$
Use the first 7 terms of the binomial expansion to establish the required inequality.
$$\begin{align}(1+1/8)^{40}&\gt\sum_{k=0}^6\frac{\binom{40}k}{8^k}\\&=1+5+\frac{195}{16}+\frac{1235}{64}+\frac{45695}{2048}+\frac{82251}{4096}+\frac{959595}{65536}\\&\gt1+5+12+19+22+20+14\\&=93\gt\frac{256}3\end{align}$$
since $93\times 3=279\gt 256$

PS: I cheated using a Python script to find the terms of the expansion to use, but all the calculations are doable by hand.
A: We can rewrite this as:
$$\left(\frac{3^4}{4^3}\right)^{17} > \left(\frac43\right)^{13}$$
The left term is $\frac{81}{64}$, which is larger than $\frac54$. So we will try to prove the more restrictive inequality:
$$\left(\frac54\right)^{17} > \left(\frac43\right)^{13}$$
This can be rearranged as:
\begin{align*}
\frac45\left(\frac54\right)^{18} &> \frac43\left(\frac43\right)^{12} \\
\left(\frac{3^25^3}{4^5}\right)^6 &> \frac53 \\
\left(\frac{1125}{1024}\right)^6 &> \frac53 \\
\end{align*}
So we can conclude using the first term of the binomial expansion of $\left(1+\frac x{1024}\right)^6 > 1+6\frac x{1024}$:
$$\left(\frac{1125}{1024}\right)^6 > 1 + 6\left(\frac{121}{1024}\right) = 1 + \frac{726}{1024} = \frac{875}{512}>\frac53$$
Which is true as $525 > 512$.
A: Another approach is to scale numbers close to one, keeping size ordering.
$\large [3^{3^4},\;4^{4^3}] 
→ [3,\;2^{128\over81}]
→ [{3\over2},\;2^{47\over81}]
→ [{9\over4},\;2^{94\over81}]
→ [{9\over8},\;2^{13\over81}]
→ [{81\over64},\;2^{26\over81}]
$
$1.26^3 = 2.000376 ≈ 2$
$81/64 = 1+ 1/4 + 1/64 > 1.26$
$26/81 = (78/81)\,/\,3 < 1/3$
If we cube both side, we have LHS > 2, RHS < 2, thus LHS is bigger.
A: $\Large {3^{3^4} \over 4^{4^3}} 
= {3^{81} \over 4^{64}} 
= {3 \over \left({256\over243}\right)^{16}}
> {3 \over (1+{13\over243})^{243\over13}}
> {3 \over e} > 1$
