For a prime $p$ and $k\ge 1$, $\phi (p^k)=p^k-p^{k-1}$

Given: A prime $p$, a natural $k\ge 1$, and an arbitrary standard representative $r$ with $0\le r<p^k$.

A class $[r{]}_{p^k}$ is a unit exactly when $\gcd (r,p^k)=1$ (For $n\ge 1$, $[a{]}_n$ is a unit if and only if $\gcd (a,n)=1$, The unit group $(\mathbb{Z}\mathrm{/}n{)}^{\times }$ and Euler's totient $\phi (n)=\mid (\mathbb{Z}\mathrm{/}n{)}^{\times }\mid$ for $n\ge 1$).

If $p\nmid r$, then $\gcd (p,r)=1$ (For a prime $p$ and any integer $a$, $\gcd (p,a)$ is $p$ when $p\mid a$ and $1$ otherwise; so $p\nmid a$ makes $p$ and $a$ coprime).

Every positive integer is a finite product of primes, and two finite prime products with the same value have the same factors up to a permutation (The fundamental theorem of arithmetic: every integer $n\ge 1$ is a product of primes, and the factorisation is unique up to order — if ${\prod }_{i<r}{p}_i={\prod }_{j<s}{q}_j$ with every $p_i$ and $q_j$ prime, then $r=s$ and $q_i=p_{\pi (i)}$ for some $\pi \in \mathrm{Sym}(r)$).

Every prime exceeds $1$ (Prime and composite integers: $p$ is prime when $p>1$ and its only positive divisors are $1$ and $p$).

The gcd of a nonzero pair is its greatest common divisor: it is a common divisor, it is at least every common divisor, and divisibility is transitive. Hence every divisor of $\gcd (a,b)$ divides both $a$ and $b$ (Common divisor, and the greatest common divisor $\gcd (a,b)$, with the convention $\gcd (0,0):=0$, Divisibility is reflexive and transitive on $\mathbb{Z}$, and is linear: if $d\mid a$ and $d\mid b$ then $d\mid ax+by$ for all integers $x,y$; also $d\mid a$ implies $d\mid ac$, $-d\mid a$ and $d\mid -a$).

The standard representatives modulo $p^k$ divisible by $p$ have cardinality $p^{k-1}$ (For a prime $p$ and $k\ge 1$, multiplication by $p$ bijects the standard representatives modulo $p^{k-1}$ with the representatives modulo $p^k$ divisible by $p$).

The standard representatives modulo $p^k$ map bijectively to $\mathbb{Z}\mathrm{/}{p}^k$, which has cardinality $p^k$, and $\phi (p^k)$ is the cardinality of its unit subset (For $n\ge 1$, every class in $\mathbb{Z}\mathrm{/}n$ has one representative $r$ with $0\le r<n$, so $\mid \mathbb{Z}\mathrm{/}n\mid =n$; while $\mathbb{Z}\mathrm{/}0$ is in bijection with $\mathbb{Z}$, The unit group $(\mathbb{Z}\mathrm{/}n{)}^{\times }$ and Euler's totient $\phi (n)=\mid (\mathbb{Z}\mathrm{/}n{)}^{\times }\mid$ for $n\ge 1$).

A finite disjoint union has cardinality equal to the sum of the cardinalities of its blocks (The sum rule: a finite disjoint union is finite with $\mid A\cup B\mid =\mid A\mid +\mid B\mid$ and $\mid {\bigcup }_{i\in I}{A}_i\mid ={\sum }_{i\in I}\mid A_i\mid$, and a sum over a finite index set splits along a partition, The cardinality $\mid A\mid$ of a finite set).