Euler Phi Function of Prime Power

Theorem

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

Then:

$\phi \left({p^n}\right) = p^n \left({1 - \dfrac 1 p}\right)$

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

When $n = 1$ the expression degenerates to:

$\phi \left({p}\right) = p - 1$

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

$\phi \left({2^k}\right) = 2^{k-1}$

From the definition of a prime number, the only number less than or equal to a prime $p$ which is not prime to $p$ is $p$ itself.

Thus it follows directly that: $\phi \left({p}\right) = p - 1$

$\Box$

$k \perp p^n \iff p \nmid k$

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

$k \in \left\{{p, 2 p, 3 p, \ldots, \left({p^{n - 1}}\right) p}\right\}$

Therefore:

$\displaystyle \phi \left({p^n}\right) = p^n - p^{n-1} = p^n \left({1 - \frac 1 p}\right)$

When $n = 1$, note that:

$\displaystyle \phi \left({p^1}\right) = p^1 \left({1 - \frac 1 p}\right) = p - 1$

demonstrating the compatibility of these definitions.

$\blacksquare$