This product contains the numbers $2$, $5$ and $10$, which multiply together to give $100$.
This means $99!$ is divisible by $100$.
The last two digits of $99!$ are therefore $00$, so the last-but-one digit is $0$.
There are plenty of other sets of numbers which together give the required factor of $100$. Some examples include $10$ and $20$, $5$ and $20$ and $2$ and $50$.