To determine the number of trailing zeros in $100!$ (100 factorial), we must understand what creates a trailing zero.

### Logical Deduction

1.  **Source of Trailing Zeros**: A trailing zero is produced by the factor $10$ in the prime factorization of the number. Since $10 = 2 \times 5$, we need to count the number of pairs of prime factors $(2, 5)$ in the prime factorization of $100!$.

2.  **Limiting Factor**: In any factorial $n!$, the number of factors of $2$ is always greater than the number of factors of $5$. This is because even numbers (multiples of 2) occur every 2 integers, while multiples of 5 occur only every 5 integers. Therefore, the number of trailing zeros is determined strictly by the number of times the prime factor **5** appears in the prime factorization of $100!$.

3.  **Legendre's Formula**: To find the exponent of a prime $p$ in $n!$, we use Legendre's Formula:
    $$ E_p(n!) = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor $$
    Where $\lfloor x \rfloor$ is the floor function (the greatest integer less than or equal to $x$).

    For this problem:
    *   $n = 100$
    *   $p = 5$

### Step-by-Step Calculation

We calculate $\left\lfloor \frac{100}{5^k} \right\rfloor$ for $k = 1, 2, 3, \dots$ until $5^k > 100$.

**Step 1: Count multiples of 5**
$$ \left\lfloor \frac{100}{5^1} \right\rfloor = \left\lfloor \frac{100}{5} \right\rfloor = 20 $$
There are 20 numbers between 1 and 100 that are divisible by 5 ($5, 10, 15, \dots, 100$). Each contributes at least one factor of 5.

**Step 2: Count multiples of 25 ($5^2$)**
$$ \left\lfloor \frac{100}{5^2} \right\rfloor = \left\lfloor \frac{100}{25} \right\rfloor = 4 $$
The numbers $25, 50, 75, 100$ are divisible by 25. Each of these contributes an *additional* factor of 5 (since $25 = 5 \times 5$). We already counted one factor of 5 in Step 1, so we add this second factor here.

**Step 3: Count multiples of 125 ($5^3$)**
$$ \left\lfloor \frac{100}{5^3} \right\rfloor = \left\lfloor \frac{100}{125} \right\rfloor = 0 $$
Since $125 > 100$, there are no multiples of 125 in the range. We stop here.

**Step 4: Sum the counts**
Total number of factors of 5:
$$ 20 + 4 = 24 $$

### Conclusion

Since there are 24 factors of 5 and more than enough factors of 2, there are 24 pairs of $(2, 5)$, resulting in 24 trailing zeros.

**Answer:** There are **24** trailing zeros in $100!$.