Example 1.6 Let the prime factorization of a natural number $n(n \geqslant 2)$ be $n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots$ $p_{k}^{\alpha_{k}}$, find the number of different positive divisors of $n$.
Solution: Any positive divisor $s$ of $n$ can be uniquely expressed as $s=p_{1}^{\beta_{1}} p_{2}^{\beta_{2} \cdots} p_{k^{k}}^{\beta^{k}}$, where $\beta_{i}$ $(i=1,2, \cdots, k)$ are integers, and $0 \leqslant \beta_{i} \leqslant \alpha_{i}$. Therefore, the positive divisors of $n$ can be constructed by sequentially determining $\beta_{1}, \beta_{2}$, $\cdots, \beta_{k}$. Since $\beta_{i}(1 \leqslant i \leqslant k)$ can be any of the numbers $0,1,2, \cdots, \alpha_{i}$, which are $\alpha_{i}+1$ numbers, the number of ways to determine $\beta_{i}$ is $\alpha_{i}+1$. Hence, by the multiplication principle, the number of different positive divisors of $n$ is