Place the numbers by their size.

Place the following numbers by their size:

$$A=2^{4^{2^{.^{.^{.^{2^{4}}}}}}},B=4^{2^{4^{.^{.^{.{4^{2}}}}}}},C=2^{2^{2^{.^{.^{.^{2^{2}}}}}}}$$

In number $C$ there are $2000$ "$2$" digits, and in numbers $B,A$ there are $500$ "$2$" and $500$ "$4$" digits. It seems to me that $C>B>A$, but I can't give a proof. Any hints?

Here is the same problem in art of problem solving. I hope that it helps.

Alternative and relatively easy solution without using logarithms. Let us set

$$h=2^{4^{2^4}}$$

$$j=4^{2^{4^2}}$$

$$k=2^{2^{2^{.^{.^{.^{2^{2}}}}}}}$$

where $h $ and $j $ only contain four numbers (two $2$ and two $4$), and $k $ contains sixteen $2$. Calculating the exponential towers starting from the top, we have

$$h=2^{4^{16}} = 2^{2^{32}} $$

$$j=4^{2^{16}} = 2^{2 \cdot 2^{16}} = 2^{ 2^{17}} $$

so that $h>j $ . Also, considering $k $ and solving its first two exponentials from the top, we get an exponential tower formed by fourteen elements where the first thirteen are $2$ and the last one is $16$. Therefore we clearly have $k>h>j$.

Now let us consider the numbers $A , B , C $ reported in the OP. Again starting to solve the exponentials from the top, we can rewrite them as

$$A=(A_1)^h $$ $$B=(B_1)^j $$ $$C=(C_1)^k $$

where $A_1$ and $B_1 $ are exponential towers similar to the initial towers $A $ and $B $, respectively, but with $500-4=496 \,\,\, $ elements instead of $500$; and where $C_1$ is an exponential tower similar to the initial $C $ but with $2000-16=1984 \,\,\, $ elements instead of $2000$.

Because $k>h>j \,\,\, $, if we could show that $C_1>A_1>B_1 \,\,\, $, then this would necessarily imply $C>A>B \,\,\, $. With this in mind, we can now repeat the same procedure above, applying it to $A_1, B_1, C_1 \,\,\, $. So we can set

$$A_1=(A_2)^h $$ $$B_1=(B_2)^j $$ $$C_1=(C_2)^k $$

where $A_2$ and $B_2 $ are again exponential towers similar to $A $ and $B$, respectively, but with $500-2\cdot 4 =492 \,\,\, $ elements instead of $500$; and where $C_2$ is an exponential tower similar to $C $ but with $2000-2 \cdot 16=1968 \,\,\, $ elements instead of $2000$. As above, because $k>h>j \,\,\, $, if we could show that $C_2>A_2>B_2$, then this would necessarily imply $C_1>A_1>B_1 \,\,\, $, and then $C>A>B \,\,\, $.

To generalize, after repeating this procedure $m $ times, we obtain two numbers $A_m$ and $B_m $ that are exponential towers similar to $A $ and $B $, respectively, but with $500-m\cdot 4 \,\,\,$ elements instead of $500$, and a third number $C_m$ that is an exponential tower similar to $C $ but with $2000-m \cdot 16 \,\,\, $ elements instead of $2000$. As above, because $k>h>j \,\,\, $, showing that $C_m>A_m>B_m \,\,\, \,\, $ necessarily implies $C_{m-1}>A_{m-1}>B_{m-1} \,\, \,\,\, $, which in turn implies $C_{m-2}>A_{m-2}>B_{m-2} \,\,\, \,\, $ and so on, until $C>A>B \,\,\, \,\, $.

So, repeating this procedure $m=124 \,\,$ times, we arrive to a point where $A_{124} $ and $B_{124} $ are exponential towers with $500-124\cdot 4 =4 \,\,\, $ elements, and $C_{124}$ is an exponential tower with $2000-124 \cdot 16=16 \,\,\, $ elements. Since $ A_{124}=h \,\, \,\,\, $, $B_{124}=j \,\, \,\,\, $, and $C_{124}=k \,\, \,\,\, $, we have $ C_{124}>A_{124}>B_{124} \,\,\, $, which necessarily implies, as shown above, $C>A>B \,\,\, $.

The problem can be re-formulated in algorithmic style as follows: $$ \begin{matrix} a_0 = 1 & b_0 = 1 & c_0 = 1 \\ & & c_1 = 2^{c_0} \\ a_1 = 4^{a_0} & b_1 = 2^{b_0} & c_2 = 2^{c_1} \\ & & c_3 = 2^{c_2} \\ a_2 = 2^{a_1} & b_2 = 4^{b_1} & c_4 = 2^{c_3} \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \\ a_k = 4^{a_{k-1}} & b_k = 2^{b_{k-1}} & c_{2k} = 2^{c_{2k-1}} \\ & & c_{2k+1} = 2^{c_{2k}} \\ a_{k+1} = 2^{a_k} & b_{k+1} = 4^{b_k} & c_{2k+2} = 2^{c_{2k+1}} \\ \cdot & \cdot & \cdot \\ \cdot & \cdot & \cdot \\ A = a_{1000} = 2^{a_{999}} & B = b_{1000} = 4^{b_{999}} & C = c_{2000} = 2^{c_{1999}} \end{matrix} $$ Now we know how to calculate these huge numbers - though very much in principle :-)

The logarithm with base $2$ may be defined formally as $\,\operatorname{lg}(x) = \ln(x)/\ln(2)$ .
But in the practice below its meaning is much simpler, as might become clear from a few examples: $\lg(2) = 1 \; ; \; \lg(4) = \lg\left(2^2\right) = 2 \; ; \; \lg(8) = \lg\left(2^3\right) = 3$ . In general : $\,\operatorname{lg}\left(2^x\right) = x$ .
We need a function to chop off the exponent from a power (tower) of two. We could have called this function "chopper", but the proper name for it in common mathematics is $2$-logarithm. Take it, or if you don't like logarithms: leave it. The whole employment of our "exponent chopper" is like in here: $2^x > 2^y \; \Longleftrightarrow \; x > y$ . So what's the problem?

We are going to employ mathematical induction. To that end, define subsequent "approximations" of $A,B,C$ as follows: $$ A_n = a_{2n} \quad ; \quad B_n = b_{2n} \quad ; \quad C_n = c_{4n} $$ As the first induction step, we calculate: $$ A_1 = a_2 = 2^4 = 16 \quad ; \quad B_1= b_2 = 4^2 = 16 \quad ; \quad C_1 = c_4 = 2^{2^{2^2}} = 65536 $$ That's not enough to establish an inequality between $A$ and $B$, so we take a second step for these: $$ A_2 = a_4 = 2^{4^{2^4}} \quad ; \quad B_2 = b_4 = 4^{2^{4^2}} = 2^{2\cdot{2^{4^2}}} $$ Logarithms base $2$ : $$ \operatorname{lg}\left(2^{4^{2^4}}\right) = 4^{2^4} = 2^{2\cdot{2^4}} \quad ; \quad \operatorname{lg}\left(4^{2^{4^2}}\right) = 2\cdot 2^{4^2} = 2^{1+4^2} $$ And again: $$ \operatorname{lg}\left(4^{2^4}\right) = 2\cdot 2^4 = 32\quad ; \quad \operatorname{lg}\left(2\cdot 2^{4^2}\right) = 1 + 4^2 = 17 $$ From $\,32 > 17\,$ it follows that $\;A_2 > B_2\,$ , namely: $\;2^{2^{32}} > 2^{2^{17}}$ .

Now assume that $\;C_n > A_n > B_n \gg 1\;$ and prove that $\;C_{n+1} > A_{n+1} > B_{n+1}$ . $$ \begin{matrix} A_n = a_{2n} & B_n = b_{2n} & C_n = c_{4n} \\ A_{n+1} = a_{2n+2} = 2^{4^{A_n}} & B_{n+1} = b_{2n+2} = 4^{2^{B_n}} & C_{n+1} = c_{4n+4} = 2^{2^{2^{2^{C_n}}}} \end{matrix} $$ Taking logarithms two times: $$ \begin{matrix} \operatorname{lg}\left(A_{n+1}\right) = 4^{A_n} & \operatorname{lg}\left(B_{n+1}\right) = 2\cdot 2^{B_n} & \operatorname{lg}\left(C_{n+1}\right) = 2^{2^{2^{C_n}}} \\ \operatorname{lg}\left(\operatorname{lg}\left(A_{n+1}\right)\right) = 2A_n & \operatorname{lg}\left(\operatorname{lg}\left(B_{n+1}\right)\right) = 1+B_n & \operatorname{lg}\left(\operatorname{lg}\left(C_{n+1}\right)\right) = 2^{2^{C_n}} \end{matrix} $$ From which it follows that $\;C_{n+1} > A_{n+1} > B_{n+1}\;$ as well. Especially: $$ C_2 > A_2 > B_2 $$ So for all $\,n \ge 2\,$ we have $\;C_n > A_n > B_n$ . Now specialize for $\,n = 500\,$ and you're done.
Conclusion : $\;C > A > B$ .   Suggestive picture (not pretending anything more):

BONUS. Somewhat more of a challenge is the following modification of the question:
in $C$ there are $1500$ numbers $2$ (instead of $2000$). Then we have: $$ A_n = a_{2n} \quad ; \quad B_n = b_{2n} \quad ; \quad C_n = c_{3n} $$ And the first induction step results in an equality $\;A_1=B_1=C_1$ : $$ A_1 = 2^4 = 16 \quad ; \quad B_1 = 4^2 = 16 \quad ; \quad C_1 = 2^{2^2} = 16 $$ A second step is needed to establish an inequality: $$ \operatorname{lg}(\operatorname{lg}(A_2)) = 32 \quad ; \quad \operatorname{lg}(\operatorname{lg}(B_2)) = 17 \\ \operatorname{lg}(\operatorname{lg}(C_2)) = \operatorname{lg}\left(\operatorname{lg}\left(2^{2^{2^{16}}}\right)\right) = \operatorname{lg}\left(2^{2^{16}}\right) = 2^{16} = 65536 $$ Finally resulting in the same as before: $\;C > A > B$ .