Let $a=p_{1}^{a_{1}} \cdots p_{n}^{a_{n}}, \quad b=q_{1}^{\beta_{1}} \cdots q_{m}^{\beta_{m}}$, where $p_{1}, \cdots, p_{n}, q_{1}, \cdots, q_{m}$ are all prime numbers and $\alpha_{1}, \cdots, \alpha_{n}, \beta_{1}, \cdots, \beta_{m}$ are all positive integers. Since $(a, b)=1$, we know that any $p_{i}(i=1, \cdots, n)$ and any $q_{i}(j=1, \cdots, m)$ cannot be equal. Therefore, by Lemma 6, we have
$$\begin{aligned} d(a b) & =d\left(p_{1}^{\alpha_{1}} \cdots p_{n}^{\alpha_{n}} q_{1}^{\beta_{1}} \cdots q_{m}^{\beta_{m}}\right) \\ & =\left(\alpha_{1}+1\right) \cdots\left(\alpha_{n}+1\right)\left(\beta_{1}+1\right) \cdots\left(\beta_{m}+1\right) \end{aligned}$$

Again, by Lemma 6, we have
$$\begin{array}{l} d(a) d(b)=d\left(p_{1}^{a_{1}} \cdots p_{n}^{\alpha_{n}}\right) d\left(q_{1}^{\beta_{1}} \cdots q_{m}^{\beta_{m}}\right) \\ \quad=\left(\alpha_{1}+1\right) \cdots\left(\alpha_{n}+1\right)\left(\beta_{1}+1\right) \cdots\left(\beta_{m}+1\right) \end{array}$$

From (46) and (47), Lemma 7 is proved.