TITLE: Is there a better way to explain why $\phi \left ( p^k \right ) = p^k - p^{k-1}$?
QUESTION [6 upvotes]: Let $p$ be a prime number and $k$ be some positive integer, and $\phi(n)$ be the Euler Phi Function, which counts the cardinality of the set of integers coprime and less than $n$. We wish to find $\phi \left (p^k \right )$.

My approach:
First have the naive answer $p^k-1$, and then take away the multiples of $p$.
There must be $p^{k-1}-1$ multiples of $p$ from $0$ to $p^k-1$. We show this below.
We have $p, 2p, 3p, ... , p^{k-1}p$, which yields $p^{k-1}$ multiples of $p$. But since the last one $p^{k-1}p$ is greater than $p^k-1$, we manually take that case away, leaving us with $p^{k-1}-1$ multiples of $p$, which is what we wanted.
Hence, $\phi \left (p^k \right ) = (p^k-1) - \left (p^{k-1} - 1 \right ) = p^k - p^{k-1}.$

But this seems like a roundabout way of acquiring the result, given how neat it is. This leads me to believe that there is a simpler explanation that yields the same result.

If it exists, what is that 'simpler explanation'?

REPLY [4 votes]: I think your explanation is the simplest. Here's another way to look at it that may provide a different kind of intuition. Factor out $p^k$ to get
$$
\phi(p^k) = p^k \left(  1 - \frac{1}{p} \right)
$$
which tells you to throw out the  fraction of the numbers less than $p^k$ that are divisible by $p$.