Let $M=lcm(2, 3, 4, \cdots, 100000)$, and $$\frac{A}{M}=1+ \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{100000}$$ Because $5^7 < 100000 < 5^8$, we know $M$ ends with 7 zeros. Multiplying $M$ on both sides of the above equation yields $$A = M + \frac{M}{2} + \frac{M}{3} + \cdots + \frac{M}{100000}$$ Consider its right side and note that all terms are multiples of 5 except $\frac{M}{5^7}$. Because $\frac{M}{5^7}$ is not divisible by $5$, therefore $A$ is not divisible by $5$. Similarly, because $2^{16} < M < 2^{17}$, all terms are multiples of 2 except $\frac{M}{2^{16}}$. Therefore $A$ is not divisible by 2. These imply $A$ and $M$ do not have 2 nor 5 as common divisors. Hence, after simplification, $q$ will still have 7 ending zeros.