[proofplan] We compute $\varphi(p^a)$ directly from its definition as the number of integers between $1$ and $p^a$ that are relatively prime to $p^a$. Since $p$ is prime, an integer in this interval fails to be coprime to $p^a$ exactly when it is divisible by $p$. We then count those multiples of $p$ by an explicit bijection and subtract from the total number of integers in the interval. [/proofplan]

[step:Identify the integers not coprime to $p^a$ as the multiples of $p$]Let \begin{align*} S := \{k \in \mathbb{N} : 1 \le k \le p^a\} \end{align*} be the set of positive integers at most $p^a$. Let \begin{align*} B := \{k \in S : \gcd(k,p^a)=1\} \end{align*} be the subset counted by Euler's totient function, and let \begin{align*} M := \{k \in S : p \mid k\} \end{align*} be the subset of multiples of $p$ in $S$. We claim that $S \setminus B = M$. First, if $k \in M$, then $p \mid k$ and $p \mid p^a$, so $p$ is a common divisor of $k$ and $p^a$. Hence $\gcd(k,p^a) \ge p > 1$, and therefore $k \notin B$. Conversely, suppose $k \in S \setminus B$. Then $\gcd(k,p^a)>1$. Since $\gcd(k,p^a)$ divides $p^a$ and $p$ is prime, every prime divisor of $\gcd(k,p^a)$ must be $p$. Thus $p \mid \gcd(k,p^a)$, and hence $p \mid k$. Therefore $k \in M$. This proves $S \setminus B = M$.[/step]

[guided]The set $B$ is exactly the set whose cardinality is $\varphi(p^a)$, so the main issue is to understand which integers are excluded from $B$. Define \begin{align*} S := \{k \in \mathbb{N} : 1 \le k \le p^a\}. \end{align*} Inside $S$, define \begin{align*} B := \{k \in S : \gcd(k,p^a)=1\} \end{align*} and \begin{align*} M := \{k \in S : p \mid k\}. \end{align*} We prove that the excluded integers $S \setminus B$ are exactly the multiples of $p$, namely $M$. First suppose $k \in M$. Then $p \mid k$ by definition of $M$. Also $p \mid p^a$ because $a \in \mathbb{N}$, so $a \ge 1$. Therefore $p$ is a common divisor of $k$ and $p^a$. Since $p$ is prime, $p>1$, and so \begin{align*} \gcd(k,p^a) \ge p > 1. \end{align*} Thus $\gcd(k,p^a) \ne 1$, so $k \notin B$. This proves $M \subseteq S \setminus B$. Conversely suppose $k \in S \setminus B$. Then $k \in S$ and $k \notin B$, so \begin{align*} \gcd(k,p^a)>1. \end{align*} The integer $\gcd(k,p^a)$ divides $p^a$. Because $p$ is prime, the prime factorisation of $p^a$ contains only the prime $p$. Hence any integer greater than $1$ dividing $p^a$ has $p$ as a prime divisor. Applying this to $\gcd(k,p^a)$ gives \begin{align*} p \mid \gcd(k,p^a). \end{align*} Since $\gcd(k,p^a)$ divides $k$, it follows that $p \mid k$. Therefore $k \in M$. This proves $S \setminus B \subseteq M$. Combining the two inclusions gives \begin{align*} S \setminus B = M. \end{align*}[/guided]

[step:Count the multiples of $p$ in the interval from $1$ to $p^a$] Define a map \begin{align*} f: \{j \in \mathbb{N} : 1 \le j \le p^{a-1}\} \to M \end{align*} by \begin{align*} f(j) := jp. \end{align*} The map is well-defined because if $1 \le j \le p^{a-1}$, then $p \le jp \le p^a$ and $p \mid jp$. The map $f$ is injective: if $f(j_1)=f(j_2)$, then $j_1p=j_2p$, and cancellation in $\mathbb{Z}$ gives $j_1=j_2$. It is surjective: if $k \in M$, then $k=jp$ for some $j \in \mathbb{N}$, and the inequalities $1 \le k \le p^a$ imply $1 \le j \le p^{a-1}$. Hence $f$ is a bijection, so \begin{align*} |M|=p^{a-1}. \end{align*} [/step]

[step:Subtract the non-coprime integers and factor the result]By definition of Euler's totient function, \begin{align*} \varphi(p^a)=|B|. \end{align*} Since $S$ has exactly $p^a$ elements and $S \setminus B=M$, we have \begin{align*} |B|=|S|-|M|. \end{align*} Using the count of $M$ from the previous step gives \begin{align*} \varphi(p^a)=p^a-p^{a-1}. \end{align*} Finally, in rational arithmetic, \begin{align*} p^a-p^{a-1}=p^a\left(1-\frac{1}{p}\right). \end{align*} Therefore \begin{align*} \varphi(p^a)=p^a-p^{a-1}=p^a\left(1-\frac{1}{p}\right). \end{align*}[/step]

[guided]The last step is purely a counting subtraction. By definition of Euler's totient function, $\varphi(p^a)=|B|$, where $B$ is the set of integers in $S$ that are coprime to $p^a$. The set $S$ contains exactly $p^a$ integers, and the previous step established that the excluded integers are exactly $M$. Hence \begin{align*} |B|=|S|-|M|. \end{align*} Substituting $|S|=p^a$ and $|M|=p^{a-1}$ gives \begin{align*} \varphi(p^a)=p^a-p^{a-1}. \end{align*} To obtain the factorized form, we rewrite the second term as $p^a/p$. This gives \begin{align*} p^a-p^{a-1}=p^a\left(1-\frac{1}{p}\right). \end{align*} Therefore \begin{align*} \varphi(p^a)=p^a-p^{a-1}=p^a\left(1-\frac{1}{p}\right). \end{align*}[/guided]