= Solution
The <Euler totient function> counts integers $1\leq a\leq n$ with $\gcd(a,n)=1$. The <Möbius function> has $\mu(1)=1$, is zero when a prime square divides $n$, and equals $(-1)^r$ when $n$ is a product of $r$ distinct primes. Consequently
$$
\sum_{d\mid n}\mu(d)=\begin{cases}1&n=1,\\0&n>1,\end{cases}
$$
because for $n>1$ the sum over squarefree divisors is $(1-1)^r$.
The <Möbius inversion formula> is \b[$g(n)=\sum_{d\mid n}\mu(d)f(n/d)$]. Substituting the divisor sum defining $f$ and collecting the coefficient of $g(e)$ gives
$$
\sum_{d\mid n}\mu(d)f(n/d)=\sum_{e\mid n}g(e)\sum_{d\mid n/e}\mu(d)=g(n).
$$
This proof uses only finite sums, so no convergence hypothesis is needed.
Classifying $1,\ldots,n$ by their greatest common divisor with $n$ gives $\sum_{d\mid n}\varphi(d)=n$: the class with greatest common divisor $n/d$ contains exactly $\varphi(d)$ integers. Applying the <Möbius inversion formula> yields
$$
\boxed{\varphi(n)=\sum_{d\mid n}\mu(d)\frac nd=n\prod_{p\mid n}(1-p^{-1}).}
$$
The <Riemann zeta function> is $\zeta(s)=\sum_{n\geq1}n^{-s}$ for $\operatorname{Re}s>1$. <Absolute convergence> justifies multiplication of <Dirichlet series>. The divisor identity for the <Möbius function> gives $\sum\mu(n)n^{-s}=1/\zeta(s)$ there. For $\operatorname{Re}s>2$ both factors below converge absolutely, and their <Dirichlet convolution> is the <Euler totient function>:
$$
\boxed{\frac{\zeta(s-1)}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\varphi(n)}{n^s}.}
$$
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