Sequential last digits?

The problem simplifies to 2^16. Without having to calculate that out fully, one can notice a pattern as we go up in powers of 2: 2, 4, 8, 16, 32, 64, 128, 256. The last digit of each power of two cycles as: 2, 4, 8, 6. Every fourth power (2^4, 2^8, 2^12, 2^16) ends with the digit 6. Therefore, 2^16, which is one of the fourth powers of 2, will also end in 6, or 2+2+2.

$\begin{aligned} 2^{2^{2^{2}}} & = 2^{2^{4}} \\ & = 2^{16} = 65536 \end{aligned}$

Hence, the last digit is $6 = 2+2+2$ .

Let's solve the more general problem: what is the last digit of any stack of powers of 2, with three or more 2's?

Observe the following table:

\begin{array}{l|*{8}{r}} k & 0&1&2&3&4&5&6&\cdots\\ \hline 2^k \bmod 10 & 1&2&4&8&6&2&4&\cdots\\ 2^k \bmod 4 &1&2&0&0&0&0&0&\cdots \end{array} Note that the second line cycles with a period of 4, when $k\ge1$ . Hence, to calculate the last digit of a power of 2, that is, $2^k \bmod 10$ , with $k\ge 1$ , it suffices to calculate $2^{k \bmod 4}$ , where $k \bmod 4$ is taken to the range $1, \ldots, 4$ (so, actually $(k-1)\bmod 4 +1$ )..

Now, if that $k$ is itself a power of 2, then apparently we need to be interested in $2^k \bmod 4$ (for another $k$ ). This is shown in the third line: it cycles with period 1 when $k\ge2$ .

Thus, any power stack of three or more 2's will end in $6=2+2+2$ .

The problem actually simplifies to 2^8 because the exponent powers rule says to multiply exponents: 2 * 2 * 2. Since 2^8 = 256. the answer is 2 + 2 + 2.

Problem: 2^2^2^2

Steps: 2^2=4 ; 4^2=16

Solution: 16^2=256 Last digit is Answer: 6= 2+2+2