jmc

Ten to the first power has two digits, $10^2$ has three digits, $10^3$ has four digits, and so on. Therefore, $10^{100}$ has 101 digits. The least number with $100$ digits is $10^{99}$, which is $10^{100}-10^{99}=10^{99}(10-1)=9\cdot 10^{99}$ less than $10^{100}$. Since $9^{100}<9\cdot 10^{99}$, $10^{100}-9^{100}$ is between $10^{99}$ and $10^{100}$. Therefore, $10^{100}-9^{100}$ has $\boxed{100}$ digits.