jmc

The pattern suggests that for a number with $n$ nines, that number squared has $n-1$ zeros. Thus, $99,999,999^2$ should have $8-1=7$ zeros. To prove this, we note that $99,999,999=10^8-1$, so $99,999,999^2=(10^8-1{)}^2=10^{16}-2\cdot 10^8+1$. Consider this last expression one term at a time. The first term, $10^{16}$, creates a number with 16 zeros and a one at the front. The second term, $2\cdot 10^8$, is a number with 8 zeros and a two at the front. The latter number is subtracted from the former one, so what is left is a string of 7 nines, then an eight, then 8 zeros. Finally, the last term changes the last zero of the number to a one. Thus, we are left with $\boxed{7}$ zeros.