Last digits of a tetration

can you explain things? it is confusing.

Actually there's a much simpler solution that let's you do with just pencil and paper in theory, thanks to the multiplicative orders:

$2019^{2019} \equiv 2019^9 \mod{10^2} \equiv 79 \mod{10^2}$

So the last 2 digits are 79. We can proceed computing the next digits in the same way:

$2019^{2019^{2019}} \equiv 2019^{79} \mod{10^3} \equiv 179 \mod{10^3}$

So the last 3 digits are 179. And so on, one can easily compute all the 12 digits without even using a calculator, with enough patience.

By the way to compute the last digits of a big power you can just compute $2019^2, 2019^4, 2019^8, 2019^{16}, 2019^{32}, ...$ , since every number can be expressed in binary, so for example if I need to compute

$2019^{79}$

I just need to multiply

$2019^{1} \times 2019^{2} \times 2019^{4} \times 2019^{8} \times 2019^{64} \mod{10^3}$ .

The question is interesting exactly because of the fact that it is actually doable without a calculator, in probably less than half an hour.

By the way I haven't put only the last 5 digits or so because otherwise one could cheat with wolframalpha very easily, by inserting

2019^2019^2019^2019^2019^2019^2019^2019^2019^2019

It would immediately tell the last 10 digits... With 12 digits anyone can still easily get the result, by putting $2019^9$ , then $2019^{79}$ , then $2019^{179}$ and so on, but at least this requires a little more work with the brain, using the multiplicative orders. Actually, one could still get immediately the answer with this command on wolframalpha:

2019^2019^2019^2019^2019^2019^2019^2019^2019^2019^2019^2019 mod 10^12

But this still requires the knowledge of fixed digits in tetrations.