How do I find the last digit of a large exponent?

Question

I am trying to find the last digit of $2^{214412412}$ using Euler's theorem. I forgot how to do this using modular arithmetic. Please, can someone explain this to me?

Answer 1

$214412412 \equiv 0 \text{ (mod 4)}$

$2$ has a cycle of $4$ in its powers. Now, can you deduce?

Answer 2

For the last digit we need to calculate $2\pmod{10}$.

Now, $2^{1} \equiv 2\pmod{10}$

$2^{2} \equiv 4\pmod{10}$

$2^{3} \equiv 8\pmod{10}$

$2^{4} \equiv 6\pmod{10}$

$2^{4n} \equiv 6^{n} \pmod{10}$ for all integers $n$

But $6^{n}\equiv 6\pmod{10}$

Thus, $2^{4n} \equiv 6\pmod{10}$

Thus, if the exponent is divisible is $4$ last digit is $6$. In your case, since the exponent has last two digits $12$ and so divisible by $4$, we conclude that the last digit of $2^{214412412}$ is $6$.