Prove that $f(p^a)=p^a=p^a$ for an arithmetic function

Question

Let $f(n)=\sum_{d|n, d>0}\phi(d)$. If p is prime, prove that $f(p^a)=p^a$. Deduce that $f(n)=n$

So far I figured out that the divisors of $p^a$ are $1, p, ..., p^a$

so $f(p^a)$=$\phi(1)+\phi(p)+...+\phi(p^a)$$=(1)+(p-1)+...+(p^{a}-1)$

I got stuck here. How can I go from here to $p^a$?

Possible duplicate of Prove that $\sum_{d|n}\phi(d)=n$ where $\phi$ is the Euler's phi function, $n,c\in\mathbb{N}$ — Dietrich Burde, Oct 21 '16 at 22:46
It's both a telescoping sum, and (aside from the initial $1$) a geometric sum. — Greg Martin, Oct 21 '16 at 23:17
@DietrichBurde: I'd say this isn't a duplicate of the other question, since this question asks about a very specific proof method, whereas the other one uses other proofs. — Greg Martin, Oct 21 '16 at 23:17

score 1 · Answer 1 · answered Oct 22 '16 at 00:58

You've made a small error in computing $\phi(p^n)$. $\phi(p^n) = p^n - p^{n-1}$, not $p^n - 1$. If $k$ is relatively prime to $p^n$, $k$ must be relatively prime to $p$. Thus, to count $\phi(p^n)$, we need to count multiples of $p$ less than or equal to $p^n$. These are $p, 2p, 3p,\dots, (p^{n-1} - 1)p, p^{n-1} p$. There are precisely $p^{n-1}$ such numbers, so $\phi(p^n) = p^n - p^{n-1}$. So the sum you want to compute is actually $$ \sum_{d\mid p^n, \, d > 0}\phi(d) = \phi(1) + \phi(p) + \dots + \phi(p^n) = 1 + (p - 1) + (p^2 - p) + \dots + (p^n - p^{n-1}). $$ Can you finish it from here?

Prove that $f(p^a)=p^a=p^a$ for an arithmetic function

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