The Last Digit

First note that $2^{2^2} = 2^4 = 16$ , and that $2^{2^3} = 2^8 = (2^4)^2 = 16^2 = 256$ . And $2^{2^4} = (2^{2^3})^2 = 256^2$ , so that the last digit of $2^{2^4}$ must be $6$ . This gives us the idea that for $n \ge 2$ , the last digit of $2^{2^n}$ is always $6$ .

We prove our idea using mathematical induction. For the basis step, we have already checked that the last digit of $2^{2^2}$ is $6$ . Suppose $n \ge 2$ and the last digit of $2^{2^n}$ is $6$ . Then the last digit of $(2^{2^n})^2$ must also be $6$ , and $(2^{2^n})^2 = 2^{2^{n+1}}$ . It follows that the last digit of $2^{2^{n+1}}$ is $6$ , and so by induction, the last digit of $2^{2^n}$ is always $6$ for $n \ge 2$ .

Therefore, the last digit of $2^{2^{2001}}$ is $6$ .