Example 7 How many positive integer factors does the number 20! have?

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Example 7 
How many positive integer factors does the number 20! have?
$\left(\alpha_{k}+1\right)$ can be calculated to yield a 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
$20!=2^{18} \cdot 3^{8} \cdot 5^{1} \cdot 7^{2} \cdot 11 \cdot 13 \cdot 17 \cdot 19$
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.