To prove when $n=1$, then since $\sum_{d \mid n} \mu(d)=\mu(1)=1$, the lemma holds.

Now suppose $n \geqslant 2$ is an integer. When $m$ is a positive integer and $m \mid n$, we use the notation $\sum_{m \mid d \mid n}$ to denote a sum over all divisors $d$ of $n$ that are divisible by $m$. In particular, when $m=1$, $\sum_{1 \mid d \mid n}$ is the same as $\sum_{d \mid n}$. Now let $p$ be a prime, then we have
$$\sum_{d \mid p} \mu(d)=1+\mu(p)=1-1=0$$

Now let $p_{1}, \cdots, p_{l}$ be $l$ distinct primes, we first prove
$$\sum_{d \mid p_{1} \cdots p_{l}} \mu(d)=0$$

When $l=1$, by (56) we know (57) holds. Now suppose $k \geqslant 2$ and (57) holds for $i=1, \cdots, k-1$, i.e.,
$$\sum_{d \mid p_{1} \cdots p_{k-1}} \mu(d)=0$$

Then by $p_{1}, \cdots, p_{k}$ being $k$ distinct primes and Lemma 13, we have
$$\begin{array}{l} \sum_{d \mid p_{1} \cdots p_{k}} \mu(d)=\sum_{d \mid p_{1} \cdots p_{k-1}} \mu(d)+\sum_{p_{k} \mid d \mid p_{1} \cdots p_{k}} \mu(d) \\ \quad=\left(1+\mu\left(p_{k}\right)\right) \sum_{d \mid p_{1} \cdots p_{k-1}} \mu(d)=0 \end{array}$$

Thus, when $l=k$, (57) also holds, and by mathematical induction, (57) holds.

Let $n=p_{1}^{a_{1}} \cdots p_{l}^{a_{l}}$, where $p_{1}, \cdots, p_{l}$ are $l$ distinct primes, and $a_{1}, \cdots, a_{l}$ are $l$ positive integers. Since $\mu(d)=0$ when $d$ is divisible by the square of a prime, by (57) we have
$$\sum_{d \mid n} \mu(d)=\sum_{d \mid p_{1} \cdots p_{l}} \mu(d)=0$$

Thus, the lemma is proved.