Finding the Unit’s Digit.

Relevant wiki: Euler's Theorem

Let the given number be $N = 2^{3^{4^a}}$ . We need to find $N \bmod 10$ . Since $\gcd(2,10) \ne 1$ , we have to consider $N \bmod 2$ and $N \bmod 5$ separately using Chinese remainder theorem .

We note that $2^{3^{4^a}} \equiv 0 \text{ (mod 2)}$ and

$\large \begin{aligned} N & \equiv 2^{3^{4^a}\color{#3D99F6} \bmod \phi(5)} \text{ (mod 5)} & \small \color{#3D99F6} \text{Since }\gcd(2,5) = 1\text{, Euler theorem applies.} \\ & \equiv 2^{3^{4^a}\color{#3D99F6} \bmod 4} \text{ (mod 5)} & \small \color{#3D99F6} \text{Euler totient function }\phi (5) = 4 \\ & \equiv 2^{3^{4^a \color{#D61F06} \bmod \phi (4)}\color{#3D99F6} \bmod 4} \text{ (mod 5)} & \small \color{#D61F06} \text{Again }\gcd(3,4) = 1\text{, Euler theorem applies.} \\ & \equiv 2^{3^{4^a \color{#D61F06} \bmod 2}\color{#3D99F6} \bmod 4} \text{ (mod 5)} & \small \color{#D61F06} \phi (4) = 2 \\ & \equiv 2^{3^{\color{#D61F06}0}\color{#3D99F6} \bmod 4} \text{ (mod 5)} \\ & \equiv 2^{\color{#3D99F6}1 \bmod 4} \text{ (mod 5)} \\ & \equiv 2 \text{ (mod 5)} \\ & \equiv 5{\color{#3D99F6}n} + 2 & \small \color{#3D99F6} \text{where }n \text{ is an integer.} \end{aligned}$

Then we have $N \equiv 5n+2 \equiv 0 \text{ (mod 2)}$ $\implies n = 0$ and $N \equiv \boxed 2 \text{ (mod 10)}$ .

Since ( $2018$ mod $10$ ) is 8 and we know that ( $2^x$ mod $10$ ) has a period of $4$ , thus the answer is $2.$