ending zeros in 100! I'm working through Hammack's Book of Proof.  Section 3.2 has an weird question, and unfortunately it's even-numbered, so there is no answer key.
"There are two 0's at the ned of 10! = 3,628,800.  Using only pencil and paper, determine how many 0's are at the end of the number 100!."
I used the special case De Polignac's formula for factorials to get tz(100!) = 24.  I believe this is the right answer. 
But the thing is,  this question (and this solution) is totally unlike everything else I've seen in the book.   This chapter is about 'Counting' (factorials, unique lists, etc).  I'm wondering if there is some other way to get the answer using tools I'd seen in text so far, and not this kind of weird excursion into number theory.  Like some intuitive way to think through the question.

The exponent of $2$ is equal to
 $$\sum_{n\ge1}\left[\frac{100}{2^n}\right]=\left[\frac{100}{2}\right]+\left[\frac{100}{2^2}\right]=75$$ and the exponent of $5$ is
$$\sum_{n\ge1}\left[\frac{100}{5^n}\right]=\left[\frac{100}{5}\right]+\left[\frac{100}{5^2}\right]=24$$ Hence the exponent of $10$ is equal to $24$. One has $$100!=10^{24}M\text{ where } (M,10)=1$$
NOTE.- It is clear that the exponent of $2$ being surely greater than the exponent of $5$ it would be enough to calculate the exponent of $5$. However I wanted to expose the general procedure without the quite particular case of $10$.
