Simple

If we substitute exponents of 2 in ascending order it becomes clear that the pattern/cycle is (2,4,8,7,5,10). Now $100\equiv4\pmod{6}$ , so we take the 4th term in the cycle. Hence, the answer is 7.

Repeatedly summing the digits of a number in this manner is equivalent to taking the number modulo 9. Thus, the answer is: $2^{100} \equiv {(2^{10}})^{10} \equiv 1024^{10} \equiv 7^{10} \equiv (7^2)^5 \equiv 49^5 \equiv 4^5 \equiv 1024 \equiv \boxed{7} \pmod 9$

Solved with a little Python:

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def digit_sum(x):
    total = 0
    for digit in str(x):
        total += int(digit)
    return total
answer = 2**100
while len(str(answer)) != 1:
    answer = digit_sum(answer)
print answer