find the remainder for $\color{purple}{TETRATION}$ such that $( {^{2021}2021}(\mod13))$ It is given that $ {^{2021}2021}$ is a tetration and I want to find the remainder when it is divided by $13$.
Briefly , I am looking for $ {^{2021}2021}\pmod{13}$
My work :
$\color{purple}{Firstly:}$ I found $2021\equiv 6\pmod{13}$
$\color{blue}{Secondly:}$ I thought that $6^{\color{red}{a}}\equiv 6^{\color{red}{b}}\pmod{13}$ should be satisfied.
$\color{orange}{Thirdly:}$ $a\equiv b\pmod{12}$ by Euler's phi function.Then , i said that $a\equiv 11$ and $b\equiv -1$
$\color{red}{Fourthly:}$ I said that $6^{11^m}\equiv 6^{(-1)^m}\pmod{13}$ and there left $2020$ superscript for tetration. Therefore , I said that it must be equal to $6^{(-1)^{2020}}\equiv6\pmod{13}$
$\therefore$ Answer is $6$
Is my solution correct , if not , can you share any trick or shortcut for finding remainders when we encounter tetrations..
$\color{red}{NOTE:}$  WHAT IS TETRATION $\rightarrow $ https://en.wikipedia.org/wiki/Tetration
$\color{red}{NOTE-2:}$ I would rather colorful answer  for the sake of @terasalisbon :)
$\color{GREEN}{THANKS...}$
 A: The only thing you need to know is that looking at a power $a^b$ modulo $k$ you can reduce the base $a$ modulo $k$ and the exponent $b$ modulo $\varphi(k)$.
Hence, when you consider the a tetration
$$
{}^{2021}2021 = \color{purple}{2021}^{\color{blue}{2021}^{\color{orange}{2021}^{\cdots^{\color{red}{2021}}}}},
$$
you can reduce the base modulo $13$ the first exponent modulo $\varphi(13)=12$, the third exponent modulo $\varphi(12)=4$, where already $2021\equiv 1 \pmod 4$.
Hence
$$
{}^{2021}2021 \equiv \color{purple}6^{\color{blue}5^{\color{orange}1^{\cdots^{\color{red}{2021}}}}} \equiv 6^5 \equiv 7776 \equiv 2 \pmod{13}.
$$
A: Tetration is a repeated raising to a given power, a "tower" of exponents. Effectively this is worked right to left (or "top down" looking at the tower analogy).
So this is a massively tall tower of exponents, but oddly enough that doesn't matter; we'll only be interested in the bottom few elements to find the answer.
You start well to find that $2021\equiv 6 \bmod 13$.
Then from Fermat's little theorem we know that $2021^{12}\equiv 1 \bmod 13$ so $2021^a$ will cycle round values on a pattern of $12$ length (at most).
Checking that, we see $(6^i){\large|_{i=1}^{12}}\equiv (6,10,8,9,2,12,7,3,5,4,11,1) \bmod 13$ in that order.
So now stepping up the tower of exponents, consider $2021^a$ where $a=2021^b$. We know that we only need to consider the values of $a\bmod 12$ from the above cycle, so that implies that we need to look at $a = 2021^b = 5^b\bmod 12$. So how does $5$ cycle under an exponent $\bmod 12$? It's easy to find by calucaltion that $5^2 = 25\equiv 1\bmod 12$, so the cycle now is only length two - effectively, we only need to know whether $b$ is odd or even.
In the tower of exponents though we know that $b = 2021^c$, which is odd. So we have our result:
$${}^{2021}2021 \equiv 6^{5^{\text{odd}}} \equiv 6^5\equiv 2 \bmod 13$$
A: $^{2021}2021 = 2021^{^{2020}2021}\pmod{13}\equiv 2021^{^{2020}2021\pmod {12}}\pmod {13}\equiv 2021^{2021^{^{2019}2021}\pmod{12}}$
$6^{5^{^{2019}2021}\pmod{12}}\pmod{13}$
Now the order of $5^2 \equiv 1 \pmod 12$ so $5^{odd} \equiv 5 \pmod{12}$ and $5^{even}\equiv 1\pmod {12}$ and $^{2019}2021$ is odd.
So $6^{5^{^{2019}2021}\pmod{12}}\pmod {13}\equiv 6^5 \equiv 6^2*6^2*6\equiv (-3)^2*6\equiv 45\equiv 2 \pmod {13}$.
