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Let $N_q(n)$ be the number of monic irreducible polynomials of polynomial degree $n$ over ${\mathbb{F}}_q$. Factoring $X^{q^n}-X$ into all such polynomials whose degrees divide $n$ givesThe Möbius inversion formula therefore gives

The Euler totient function counts integers $1\le a\le n$ 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. Consequentlybecause for $n>1$ the sum over squarefree divisors is $(1-1{)}^r$.

The Möbius inversion formula is $g(n)={\sum }_{d\mid n}\mu (d)f(n/d)$. Substituting the divisor sum defining $f$ and collecting the coefficient of $g(e)$ givesThis proof uses only finite sums, so no convergence hypothesis is needed.

Classifying $1,\dots ,n$ by their greatest common divisor with $n$ gives ${\sum }_{d\mid n}\phi (d)=n$: the class with greatest common divisor $n/d$ contains exactly $\phi (d)$ integers. Applying the Möbius inversion formula yieldsThe Riemann zeta function is $\zeta (s)={\sum }_{n\ge 1}{n}^{-s}$ for $\mathrm{Re}s>1$. 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 $\mathrm{Re}s>2$ both factors below converge absolutely, and their Dirichlet convolution is the Euler totient function:

An arithmetic function $f$ is multiplicative if $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are coprime integers. If $g(n)={\sum }_{d\mid n}f(d)$ and $(m,n)=1$, every positive integer divisor of $mn$ is uniquely $d_1{d}_2$ with $d_1\mid m$ and $d_2\mid n$. Henceso $g$ is multiplicative.

The Möbius function is defined by $\mu (1)=1$, $\mu (n)=0$ if the square of a prime number divides $n$, and $\mu (n)=(-1{)}^r$ if $n$ is a product of $r$ distinct prime numbers. Coprime integers have disjoint sets of prime factors, so this definition immediately gives $\mu (mn)=\mu (m)\mu (n)$ when $(m,n)=1$.

For $n=p_1^{a_1}\cdots p_r^{a_r}>1$, only the square-free divisors contribute, and thereforeFor $n=1$ the sum is $1$. Thus $1\ast \mu =\epsilon$ in the language of Dirichlet convolution. Convolving $g=1\ast f$ with $\mu$ gives the Möbius inversion formula

If $f(n)=n$, its divisor sum is the sum-of-divisors function. Conversely, if $g(n)=n$, Möbius inversion givesusing the standard divisor identity for the Euler totient function. Therefore the two requested answers are

An arithmetic function $f$ is multiplicative when $f(1)=1$ and $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are coprime. If $m$ and $n$ are coprime, every divisor $d$ of $mn$ has a unique factorization $d=d_1{d}_2$ with $d_1\mid m$ and $d_2\mid n$. Hence the Dirichlet convolution of multiplicative functions satisfiesand $(f\star g)(1)=1$, so it is multiplicative.

Let $\epsilon$ be the Identity for Dirichlet convolution, with $\epsilon (1)=1$ and $\epsilon (n)=0$ for $n>1$. The Möbius function satisfies $\mu \star 1=\epsilon$. Therefore, using commutativity and associativity,This is the Möbius inversion formula.