Prove that $\varphi(p^k)=p^k-p^{k-1}$ for prime $p$

Question

Let $p$ be a positive prime, and $k\in\mathbb{N}$. Prove that $\varphi(p^k)=p^k-p^{k-1}$ (Euler's totient)

I suppose maybe induction should be involved. But I'm having difficulty relating $p^k$'s prime factors to $p^{k-1}$ prime factors exactly. I suppose that for example, $p^2$'s factors are those that are multiples of $p$'s factors. But how do I know that the numbers bigger than $p$ that are factors of $p^2$ add with the factors of $p$ to equal $p$? Sorry, does that even make sense?

Answer 1

Hint:

Make a list of all the numbers between $1$ and $p^k$. $$1,2,3,4,\ldots, p^k$$

• How many numbers are there on this list?

  There are $p^k$ numbers on the list.

• How many numbers have a common factor with $p^k$?

  Since $p$ is a prime another number will have a common factor with $p^k$ if and only if $p$ divides it. These numbers are $$p\cdot \color{green}{1}, p\cdot \color{green}{2}, p\cdot \color{green}{3}, \ldots, p\cdot \color{green}{p^{k-1}}$$ There are $p^{k-1}$ such numbers.

• How many numbers does not have a common factor with $p^k$? (This is $\phi(p^k)$)

Answer 2

Notice that for the definition of prime number, a number $n$ is coprime with $p^k$ if and only if $p$ doesn't divide $n$. Since there are $\frac{p^k}{p}=p^{k-1}$ multiples of $p$ then:

$$\varphi(p^k)=p^k-p^{k-1}$$