What would the result be when this method is applied to Graham's number?
(This is very similar to an earlier question, "Logarithms of logarithms of Graham's number, is the result ever handy?".)
Answer: The method cannot be applied to Graham's number ($G$), because the number of iterations required would be too large to have its digits written on any card.
By the definition of $G$ and the properties of Knuth arrows,
$$G \ = \ 3\uparrow^{g_{63}} 3 = 3\uparrow^{g_{63}-1}(3\uparrow^{g_{63}-1} 3) \ \ggg \ 3\uparrow^{2}(3\uparrow^{g_{63}-1} 3) \ \gt \ 10\uparrow^{2}\left( \frac{3\uparrow^{g_{63}-1} 3}{9}\right)$$
where $g_{63}$ is a number with an incomprehensibly large number of decimal digits. (The rightmost inequality is by application of a simple change-of-base inequality.)
In the present problem, the number of full pages occupied by the digits of a number x is
$$f(x) \ = \ \left\lfloor \frac{\lfloor \log_{10} x \rfloor +1}{4000} \right\rfloor $$
and the number to be finally written on the card is the least number of iterations, say $n$, such that
$$f^n(G) \ \le \ 25000$$
However, it's easily verified that
$$ f(x) \ \gt \ \log_{10} \log_{10} x \quad (x \ge 10^{10^{10}}) $$
so for any number of iterations $k$ (until the tower is reduced to less than $10^{10^{10}}$),
$$f^k(G) \ \gt \ (\log_{10})^{2k}(G) \ \ggg \ 10\uparrow^{2}\left( \frac{3\uparrow^{g_{63}-1} 3}{9}-2k\right)\tag{*} $$
Therefore, long before the final iteration is reached, the number of iterations will become utterly "unwritable". For example, even for $k$ iterations, where $k$ is the "unwritable" number
$$k = 3\uparrow^{g_{63}-2} 3$$
the intermediate result, $f^k(G)$, is still enormously larger than a tower of $10$s of height
$$\frac{3\uparrow^{g_{63}-1} 3}{9}-2(3\uparrow^{g_{63}-2} 3) \ \ggg \ 3\uparrow^{g_{63}-2} 3$$
Note that $3\uparrow^{g_{63}-2} 3$ has only two less arrows than $G$ itself!
NB: Graham's number is extreme overkill for this conclusion. For example, the cartoon's method cannot be applied to any number $x$ greater than or equal to, say, $10\uparrow\uparrow{(10\uparrow\uparrow5)}$, because the lefthand inequality in (*) above (with $x$ replacing $G$) will then show that $f^k(x) \ \gt \ 10\uparrow\uparrow4$ even for $k > 10\uparrow\uparrow4$ iterations. ($10\uparrow\uparrow4$ is "unwritable" in the sense of having more decimal digits than will fit in the observable universe, assuming at least one Planck volume per digit.)
