A small tower

$2^1=2\\2^2=4\\2^3=8\\2^4=16\\2^5=32\\2^6=64\\2^7=128\\2^8=256\\2^9=512\\\vdots\quad \vdots\quad \vdots$

We can see, changing exponent $4$ , the last digit remains same. Again if exponent is divisible by 4 than the last digit come $6$ .

Here, the $2$ 's exponent $\LARGE{ 4^{6^{8^{10}}}}$ is divisible by $4$ .

So, it's last digit is $6$

Relevant wiki: Euler's Theorem

Easy solution:

Note that $2^{4^a} = 2^4 \cdot 2^4 \cdot 2^4 \cdots 2^4 = 16 \cdot 16 \cdot 16 \cdots 16$ . We know that the product of two numbers end with 6 also ends with 6. Therefore, the last digit of a number of the form $2^{4^a}$ , where $a$ is a natural number, is $\boxed{6}$ .

Chinese remainder theorem

Since $\gcd(2,10) \ne 1$ , we have to consider the prime factor 2 and 5 separately using Chinese remainder theorem as follows.

$\begin{aligned} 2^{4^a} & \equiv 0 \text{ (mod 2)} \implies 2^{4^a} \equiv 2n \end{aligned}$

Now consider:

$\begin{aligned} 2^{4^a} & \equiv 2^{\color{#3D99F6}4^a \bmod \phi(5)} \text{ (mod 5)} & \small \color{#3D99F6} \text{Since }\gcd(2,5) = 1 \text{, Euler's theorem applies.} \\ & \equiv 2^{\color{#3D99F6}4^a \bmod 4} \text{ (mod 5)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (5) = 4 \\ & \equiv 2^0 \text{ (mod 5)} \\ \implies 2n & \equiv 1 \text{ (mod 5)} \\ \implies n & \equiv 3 \\ \implies 2^{4^a} & \equiv 2n \equiv \boxed{6} \text{ (mod 5)} \end{aligned}$

Though others solutions are nice but for me, there can be much easier solution than those. $2≡2mod 10$

→ $2^4≡6 mod 10$

→ $(2^4)^x ≡6^x mod 10$

→ $2^{4x} ≡ 6 mod 10$ (since 6^n ends with 6 for every natural number n)

So last digit of $2^{4x}$ is 6 where x is any natural number.

Now clearly, $2^{4^{6^{8^{10}}}}$ $= 2^{4^n} = 2^{4x}$ hence its last digit must be $6$

\large \log_2 2^{4^{6^{8^{10}}}}=4.6.8.10.\log_2 2=1920\log_2 2 = 1920\\ \large 1920\mod{4}=0\\  2^{1}=2;          2^{2}=4;          2^{3} =8;          2^{4}=16;\\ 2^{5}=32;        2^{6}=64;        2^{7}=128;      2^{8}=256;\\           \vdots\quad                   \vdots\quad                 \vdots\quad                 \vdots\quad\\ 2^{1917}=..2;    2^{1918}=..4;    2^{1919}=..8;    2^{1920}=..6;

The last digit of 2^4 is 6. No matter how many numbers you multiply, as long as 6 is the last digits of all the numbers, the last digit of the product must be 6