Euler Phi Function of Prime Power

Theorem
Let $p^n$ be a prime power for some prime number $p > 1$.

Then:
 * $\map \phi {p^n} = p^n \paren {1 - \dfrac 1 p} = \paren {p - 1} p^{n - 1}$

where $\phi: \Z_{>0} \to \Z_{>0}$ is the Euler $\phi$ function.

Corollary
When $p = 2$, the formula is exceptionally simple:

Proof
From Euler Phi Function of Prime:
 * $\map \phi p = p - 1$

From Prime not Divisor implies Coprime:
 * $k \perp p^n \iff p \nmid k$

There are $p^{n - 1}$ integers $k$ such that $1 \le k \le p^n$ which are divisible by $p$:


 * $k \in \set {p, 2 p, 3 p, \ldots, \paren {p^{n - 1} } p}$

Therefore:
 * $\map \phi {p^n} = p^n - p^{n - 1} = p^n \paren {1 - \dfrac 1 p} = \paren {p - 1} p^{n - 1}$

Also see

 * Euler Phi Function of Prime: When $n = 1$ the expression degenerates to $\map \phi p = p - 1$.