Math is fun

Last digit of 2 follows a recurring cycle . Since any power of 4 is divisible by 4, last digit of

$\huge{{2}^{4^n}}$

will always be 6 for integral $n$ .

Hence the answer is $6 + 1 = 7$

If we work modulo 5 (and modulo $\phi(5)=4$ in the exponent), the given number $n$ is $n\equiv 2^0+1=2 \pmod{5}$ . Since $n$ is odd, we have $n\equiv \boxed{7} \pmod{10}$ .

2^4=16. 16^n will always end with 6 so 6+1=7