It should be pretty obvious why $10!$ and all higher factorial must all have at least one zero at the end: they're all divisible by $10$.
$$10! = \mathbf{10} \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$$
If you think about it a bit more, it's also pretty obvious that $20!$ and any factorials above it must haveat least two zeros at the end (because they're divisible by $10 \times 20$) and that $30!$ and above must have at least three zeros (because they're divisible by $10 \times 20 \times 30$) and so on.
Actually, this "rule" underestimates the number of zeros at the end of factorials by about a factor of $2$. Why? Because $2 \times 5 = 10$, so $5! = 5 \times 4 \times 3 \times 2 \times 1$ already has one zero at the end, and every further multiple of $5$ adds yet another zero (there being plenty of even numbers to provide the multiples of $2$ needed to make up $10$). So $15!$ has three zeros at the end, not just one, and $20!$ actually has four, not two.
Also, $25!$ actually gains two extra zeros from being a multiple of $25 = 5 \times 5$, for a total of six. The same happens at $50!$ and $100!$, and $125!$ actually has three more zeros at the end than $124!$, because $125 = 5 \times 5 \times 5$.
So, looking at your example, we don't actually need to calculate $85!$ to tell that it has twenty zeros at the end: one each from $5$, $10$, $15$, $20$, $25$, $30$, $35$, $40$, $45$, $50$, $55$, $60$, $65$, $70$, $75$, $80$ and $85$, and one extra each from $25$, $50$ and $75$.