Example 7 How many positive integer factors does 20! have?
Analyze writing 20! in its standard factorization form $n=\beta_{1}^{a_{1}} \beta_{2}^{a_{2}} \cdots \beta_{k}^{q_{k}}$, and then using $r(n)=\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots$ $\left(\alpha_{k}+1\right)$ to calculate the result.
Solution Since the prime numbers less than 20 are $2,3,5,7,11,13,17,19$, in the standard factorization of 20!, the highest power of 2 $=\left[\frac{20}{2}\right]+\left[\frac{20}{4}\right]+\left[\frac{20}{8}\right]+\left[\frac{20}{16}\right]=18$,
the highest power of 3 $=\left[\frac{20}{3}\right]+\left[\frac{20}{9}\right]=8$,
the highest power of 5 $=\left[\frac{20}{5}\right]=4$,
the highest powers of $7,11,13,17,19$ are $2,1,1,1,1$ respectively.
Thus, the standard factorization of 20! is
$$20!=2^{18} \cdot 3^{8} \cdot 5^{4} \cdot 7^{2} \cdot 11 \cdot 13 \cdot 17 \cdot 19$$

Therefore, the number of positive divisors $r(20!)$ of 20! is:
$$\begin{aligned}
r(20!) & =(18+1)(8+1)(4+1)(2+1)(1+1)(1+1)(1+1)(1+1) \\
& =19 \cdot 9 \cdot 5 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \\
& =41040
\end{aligned}$$

Thus, the number of positive divisors of $20!$ is 41040.