Prove that for Re(s)>1
$$\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}$$
Where $\mu(n)$ is the Möbius function defined by:
$\mu(n)=1, \mbox{if }n=1$
$\mu(n)=(-1)^k, \mbox{if }n=p_1,p_2,...,p_k$ and $p_j$ are distinct primes
$\mu(n)=0, otherwise$
Hint: use the Euler product formula.
I started trying to use the Euler product formula for $\zeta(s)$ but no succes. How do I get this result?
Euler Product$$\zeta(s)=\prod_{p}\frac{1}{1-p^{-s}}$$
$$\frac{1}{\zeta(s)}=\prod_{p}(1-p^{-s})=(1-p_1^{-s})(1-p_2^{-s})(1-p_3^{-s})...$$
Due to the fundamental theorem of arithmetic, this product can be re-written as a sum over the integers. By foiling out the primes, each term in the sum can be seen to be made up of a product of different combinations of primes, and the sign of the term is based on the amount of prime factors it contains.
So $$\frac{1}{\zeta(s)}=\sum_{n=1}^\infty\frac{\mu(n)}{n^s}$$