Proof

Let $n\ge 1$ be a natural number.
For $n=1,$ the sum of the Möbius function over the divisors of $n$ equals $\sum_{d\mid 1}\mu(d)=\mu(1)=1,$ as required.
For $n > 1$ with the factorization $n=p_1^{e_1}\cdots p_r^{e_r},$ we have $\mu(d)=0$ for non square-free divisors $d\mid n.$
In the remaining sum, there are exactly $\binom rk$ square-free divisors $d$ with exactly $k$ different prime factors.
By the definition of the Möbius function, it follows $\begin{array}{rcl} \sum_{d\mid n}\mu(d)&=&\sum_{d\mid n,\text{ square-free}}\mu(d)=1+\binom{r}{1}(-1)+\binom{r}{2}+\cdots+\binom{r}{2}(-1)^r\\ &=&\sum_{k=0}^r\binom rk(-1)^k. \end{array}$
The last sum equals by the binomial theorem $(1-1)^r=0.$
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Scheid Harald: "Zahlentheorie", Spektrum Akademischer Verlag, 2003, 3rd Edition
Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927