Last Digit of a Fricking Power Tower

Let the given number be $3^{4^N}$ . Then we have:

$3^{4^N} \equiv 3^{2^{2N}} \equiv 3^{2\cdot2^{2N-1}} \equiv 9^{2^{2N-1}} \equiv (10-1)^{2^{2N-1}} \equiv (-1)^{2^{2N-1}} \equiv \boxed 1 \text{ (mod 10)}$

If we take the powers of $3$ , we have: $3^1\equiv3\pmod{10}$

$3^2\equiv9\pmod{10}$

$3^3\equiv7\pmod{10}$

$3^4\equiv1\pmod{10}$

$3^5\equiv3\pmod{10}$

$3^6\equiv9\pmod{10}$

$\ldots$

As you can see, if $n \mod 4\equiv0$ , $3^n\mod10$ will always be $1$ .

And because $4^{5^{\ldots^{999999999999999}}}$ is definitely divisible by $4$ , $3^{4^{5^{\ldots^{999999999999999}}}}\mod10\equiv1$

Therefore the answer is $\color{#20A900}\boxed{1}$