Let $f(n)$ be the number of ones that occur in the decimal representations of all the numbers from 1 to $n$.  For example, this gives $f(8)=1$, $f(9)=1$, $f(10)=2$, $f(11)=4$, and $f(12)=5$.  Determine the value of $f(10^{100})$.
1. **Claim**: 
   \[
   f(10^k) = 1\binom{k}{1}9^{k-1} + 2\binom{k}{2}9^{k-2} + \cdots + k\binom{k}{k}9^{k-k} + 1
   \]

2. **Proof**:
   - Consider all numbers less than \(10^k\). These numbers have \(k\) or fewer digits.
   - For numbers with exactly \(i\) ones, there are \(\binom{k}{i}\) ways to choose the positions of the ones.
   - The remaining \(k-i\) positions can be filled with any of the 9 digits that are not one, giving \(9^{k-i}\) combinations.
   - Therefore, there are \(\binom{k}{i}9^{k-i}\) numbers with exactly \(i\) ones.
   - Each of these numbers contributes \(i\) ones to the function \(f(n)\).
   - Summing over all possible \(i\) from 1 to \(k\) and adding one for \(10^k\) itself, we get:
     \[
     f(10^k) = \sum_{i=1}^{k} i \binom{k}{i} 9^{k-i} + 1
     \]

3. **Simplification**:
   - Using the identity \(i \binom{k}{i} = k \binom{k-1}{i-1}\), we can rewrite the sum:
     \[
     \sum_{i=1}^{k} i \binom{k}{i} 9^{k-i} = k \sum_{i=1}^{k} \binom{k-1}{i-1} 9^{k-i}
     \]
   - Recognizing that \(\binom{k-1}{i-1}\) is just a shifted binomial coefficient, we can rewrite the sum:
     \[
     k \sum_{i=0}^{k-1} \binom{k-1}{i} 9^{k-1-i}
     \]
   - By the binomial theorem, \(\sum_{i=0}^{k-1} \binom{k-1}{i} 9^{k-1-i} = (1+9)^{k-1} = 10^{k-1}\).
   - Therefore:
     \[
     k \cdot 10^{k-1}
     \]
   - Adding the one for \(10^k\) itself, we get:
     \[
     f(10^k) = k \cdot 10^{k-1} + 1
     \]

4. **Conclusion**:
   - For \(k = 100\):
     \[
     f(10^{100}) = 100 \cdot 10^{99} + 1 = 10^{101} + 1
     \]

The final answer is \(\boxed{10^{101} + 1}\)