Show that $\phi(n^e) = n^{e-1}\phi(n)$ where $\phi$ is the totient function

Question

So i know that $\phi(ab) = \phi(a)\phi(b)$ if $gcd(a,b)=1$ and if p is prime then $\phi(p^e) = p^e - p^{e-1}$.

I was thinking of splitting $n^e$ into product of things so that I can use the multiplicative property but im not sure how i can ensure the split's gcd is still 1.

Answer 1

Show the value of $\frac{\phi(n)}{n}$ depends only on the distinct prime factors of $n.$

So if $m_1,m_2$ have the same set of prime factors, then $$\frac{\phi(m_1)}{m_1}=\frac{\phi(m_2)}{m_2}.$$

$n$ and $n^e$ have the same prime factors.

So: $$\frac{\phi(n)}{n}=\frac{\phi(n^e)}{n^e}$$ Multiply both sides by $n^e$ to get your result.

Answer 2

HINT: The set of positive integers $M < n^e$ relatively prime to $n^e$ are precisely those integers $M$ of the following form: $M = jn + k_M$, where $j \in \{0,1,2, \ldots, n^{e-1}-1\}$, and $k_M \in \{1,2,\ldots, n-1\}$ satisfying $(k_M,n)=1$. So $\phi(n)$ choices for $k_M$, and $n^{e-1}$ choices for $j$....

If $n$ is prime, then $\phi(n)=n-1$, and $n^{e-1}(n-1)=n^e-n^{e-1}$.

Answer 3

Here's a different way to exploit multiplicativity. As both sides of the equation are multiplicative functions sending $n=1$ to $1$, you need only check equality in the case $n=p^k$ with $p\in\Bbb P,\,k\ge1$, which is trivial.