Totient of a prime power

Statement

Let $p$ be a prime number and let $k$ be a positive integer. Then $\varphi(p^k) = p^{k-1} (p - 1) = p^k - p^{k-1}$, where $\varphi$ is Euler's totient function.

Remarks

The proof: an integer is coprime to $p^k$ if and only if $p$ does not divide it. Among $1, \dots, p^k$, the multiples of $p$ are $p, 2p, \dots, p^{k-1} \cdot p$, so there are $p^{k-1}$ of them. For example, $\varphi(8) = 4$ and $\varphi(27) = 18$. With the multiplicativity of the totient function, this formula gives $\varphi(n)$ from the prime factorization of $n$.

Explanations

L1 human

Among the numbers $1$ to $p^k$, exactly one in $p$ is a multiple of $p$. Remove these numbers. The fraction $1 - 1/p$ of the numbers stays, so $\varphi(p^k) = p^k (1 - 1/p)$.