Answer:
Consider the sequence (1, 1, 2, 3, 4, 5, 6, 6, ...) of the number of digits in the powers (1, 7, 49, 343, 2401, 16807, 117649, 823543, ...) of 7.
Check: 7²⁰⁰ = (7²)¹⁰⁰ = 49¹⁰⁰ > 10¹⁰⁰
Since 10¹⁰⁰ has 101 digits and 7²⁰⁰ is even larger, 7²⁰⁰ must have more than 100 digits. In particular, the 201st term of (1, 1, 2, 3, 4, 5, 6, 6, ...) tells the number of digits in 7²⁰⁰, and this 201st term is greater than 100.
So, the sequence (1, 1, 2, 3, 4, 5, 6, 6, ...) starts below 100 and eventually exceeds 100. Next, we will show that the sequence (1, 1, 2, 3, 4, 5, 6, 6, ...) can never skip over any natural number.
Specifically, if n is a term in (1, 1, 2, 3, 4, 5, 6, 6, ...) then the next term cannot be greater than or equal to n + 2, skipping n + 1. Indeed, if 7ᵏ has n digits, then 10 × 7ᵏ has n + 1 digits, and since
7ᵏ < 7ᵏ⁺¹ = 7(7ᵏ) < 10(7ᵏ)
thereby, 7ᵏ⁺¹ has either n or n + 1 digits. Thus, each subsequent term of the sequence (1, 1, 2, 3, 4, 5, 6, 6, ...) either remains constant or increases by 1.
Finally, because (1, 1, 2, 3, 4, 5, 6, 6, ...) starts below 100 and increases to a number over 100, never skipping over any natural number, the sequence must include 100. This shows that there must exist a power of 7 with exactly 100 digits.
