Find $2^{3^{100}}$ (mod 5) and its last digit The question is to find $2^{3^{100}} \pmod 5$ and its last digit.
I think we have to find $2 \pmod 5$ and $3^{100}\pmod 5$ separately, right?
$$2 = 2 \pmod 5$$
$$3^4 = 1 \pmod 5$$
$$3^{100} = 1 \pmod 5$$
$$2^{3^{100}} = 2^1 = 2 \pmod 5$$
Is this solution correct? And to find the last digit do we just solve modulo $10$?

Yes, you solve modulo $10$ for the last digit. Note that the number is even, so its last digit is taken from $2, 5, 6, 8$--it's not divisible by $5$ so the last digit cannot be $0$. Your original solution is a bit off, though $3^4\equiv 1\mod 5$, but in the exponent it's every $4$ which gives $1$, so we want $3^2\equiv 1\mod 4$. Then $3^{100}\equiv 1\mod 4$. So
$$2^{3^{100}}=2^{4k+1}\equiv (16)^k\cdot 2\mod 5$$
which gives the same result, but by sounder methods. Since it is $2$ mod $5$ as you have said, the only choices are $2$ and $7$. $7$ is odd, so it must be $2$.
