\(101^{100} - 1 = (1 + 100)^{100} - 1 = \binom{100}{1}100 + \binom{100}{2}100^2 + \binom{100}{3}100^3 + \dots\)

\(= 10^4 + 4950 \times 10^4 + \binom{100}{3} \times 10^6 + \dots\)

\(= 10^4\left[1 + 4950 + 100(\dots)\right] = 10^4 \times (\text{a number ending in } 51)\)

The bracket ends in 1, so it is not divisible by 10.

So the highest power of 10 dividing the number is \(10^4\), and \(m = 4\).

Answer: (D) 4