For a prime $p$ and $k\ge1$, $\varphi(p^k)=p^k-p^{k-1}$

Given: A prime $p$, a natural $k\ge1$, 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 $n\ge1$, $[a]_n$ is a unit if and only if $\gcd(a,n)=1$, The unit group $(\mathbb{Z}/n)^\times$ and Euler's totient $\varphi(n)=\lvert(\mathbb{Z}/n)^\times\rvert$ for $n\ge1$).

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 $\pi \in \operatorname{Sym}(r)$).

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 $k\ge1$, 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 $\mathbb Z/p^k$, which has cardinality $p^k$, and $\varphi(p^k)$ is the cardinality of its unit subset (For $n\ge 1$, every class in $\mathbb{Z}/n$ has one representative $r$ with $0\le r<n$, so $\lvert\mathbb{Z}/n\rvert=n$; while $\mathbb{Z}/0$ is in bijection with $\mathbb{Z}$, The unit group $(\mathbb{Z}/n)^\times$ and Euler's totient $\varphi(n)=\lvert(\mathbb{Z}/n)^\times\rvert$ for $n\ge1$).

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 $\lvert A \cup B\rvert = \lvert A\rvert + \lvert B\rvert$ and $\lvert\bigcup_{i \in I} A_i\rvert = \sum_{i \in I}\lvert A_i\rvert$, and a sum over a finite index set splits along a partition, The cardinality $\lvert A\rvert$ of a finite set).