Euler Phi Function of Prime Power

Theorem
Let $p$ be a prime number, $p > 1$.

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

where $\phi: \Z^*_+ \to \Z^*_+$ is the Euler $\phi$ function.

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


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

Proof

 * $\phi \left({p}\right) = p - 1$ follows directly from the definition of prime number.

The only number less than or equal to a prime $p$ which is not prime to $p$ is $p$ itself.


 * Next, note that $k \perp p^n \iff p \nmid k$, which follows from Prime Not Divisor then Coprime.

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

Proof of Corollary
This follows directly, as $\displaystyle 1 - \frac 1 2 = \frac {2 - 1} 2 = \frac 1 2$.

Thus:
 * $\phi \left({2^k}\right) = \left({\dfrac 1 2}\right) 2^k = 2^{k-1}$