(EDITED)
$$2^{100} = \left ( 2^{10} \right )^{10} = 1024^{10} = \left ( 1000 + 24 \right )^{10} = 1000^{10} + \sum_{i = 0}^{9} c_i 1000^{i} 24^{10 - i}$$
Where the $c_i$ are the binomial coefficients for this binomial term:
$$c_i = \left ( 1, 10, 45, 120, 210, 252, 210, 120, 45, 10 \right )$$
Now observe that for $0 \leq i \leq 9$ we have that:
$$c 1000^i 24^{10 - i} < \frac{9}{10} 1000^{10} ~ ~ \iff ~ ~ c 24^{10 - i} < \frac{9}{10} 1000^{10 - i} ~ ~ \iff ~ ~ c < \frac{9}{10} 40^{10 - i} < \frac{9}{10} \left ( \frac{1000}{24} \right )^{10 - i}$$
It is easy to see that this inequality holds true for all the binomial coefficients:
$$c_0 = 1 < \frac{9}{10} 40^{10}$$
$$c_1 = 10 < \frac{9}{10} 40^9$$
$$\cdots$$
$$c_7 = 120 < \frac{9}{10} 40^3$$
$$c_8 = 45 < \frac{9}{10} 40^2$$
$$c_9 = 10 < \frac{9}{10} 40^1$$
So we can say that:
$$c_i 1000^i 24^{10 - i} < \frac{9}{10} 1000^{10} ~ ~ ~ \left ( 0 \leq i \leq 9 \right )$$
And so:
$$\sum_{i = 0}^{9} c_i 1000^i 24^{10 - i} < 10 \cdot \frac{9}{10} \cdot 1000^{10} = 9 \cdot 1000^{10}$$
We therefore conclude that:
$$1000^{10} = 10^{30} < 2^{100} < 1000^{10} + 9 \cdot 1000^{10} = 10 \cdot 1000^{10} = 10^{31}$$
And so $2^{100}$ has $31$ digits.