Show that $$\frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^{s}}.$$
I'm trying to show the above and we haven't covered Euler's product formula yet so I don't think we can use that in our proof. I think that maybe we could use Dirichlet convolution in some way but that's defined for sums over the divisors of n, which the above are not so I'm stuck. Any hints on the way to proceed without Euler's product formula would be appreciated, thanks!
Here is a more general answer. Let $\Bbb N$ here denote the set of positive integers, and suppose that $f:\Bbb N\to\Bbb C$ is an arithmetic function such that $f(1)\neq 0$. Define $$F(s)=\sum_{n=1}^\infty\frac{f(n)}{n^s}$$ where $s\in S$ and $S\subseteq\Bbb C$ is the set of values $s$ s.t. $F(s)$ absolutely converges.
Then for all $s\in S$, $$\frac{1}{F(s)}=\sum_{n=1}^\infty\frac{\hat{f}(n)}{n^s},$$ where $\hat{f}$ is the Dirichlet inverse of $f$. That is, $\hat{f}:\Bbb N\to\Bbb C$ is the unique arithmetic function s.t. $$\hat{f}*f=\epsilon=f*\hat{f},$$ where $*$ is the Dirichlet convolution and $\epsilon$ is the identity function for the Dirichlet convolution. This is because $$\left(\sum_{n=1}^\infty\frac{f(n)}{n^s}\right)\left(\sum_{n=1}^\infty\frac{\hat{f}(n)}{n^s}\right)=\sum_{n=1}^\infty \frac{(f*\hat{f})(n)}{n^s}=\sum_{n=1}^\infty\frac{\epsilon(n)}{n^s}=1.$$
In the OP's problem where $F=\zeta$, $f$ is the constant function $\texttt{1}$. Since $\mu*\texttt{1}=\epsilon$, $\mu=\hat{\texttt{1}}$. The claim follows.