TITLE: Is my proof clear/correct?
QUESTION [5 upvotes]: Proof that $\displaystyle\sum_{d|n}\phi (d)=n:$
Lemma: $\phi(a)\phi(b)=\phi(ab)$ if $(a,b)=1$. 
Proof (edited):  $\phi(n)$ is the number of generators of $C_n$. In particular the number of generators of $C_{ab}$ is $\phi(ab)$. Note that
if $(a,b)=1$ then $C_{ab}\cong C_a\times C_b$.
For any generators $\alpha,\beta$ of $C_a,C_b$ respectively, $(\alpha,\beta)$ generates $C_{ab}$, since: $$\text{ord}(\alpha,\beta)=\text{lcm}(\text{ord} \,\alpha,\text{ord}\,\beta)=\text{lcm}(a,b)=ab$$ The number of pairs of generators $(\alpha,\beta)$ is $\phi(a)\phi(b)$ which proves the claim.
Note (in respone to DonAntonio's comment): WLOG take $x$ where $\langle x\rangle$ is a proper subgroup of $C_a\;($so $\text{ord}\,x<a)$ and $y\in C_b$. By Lagrange $\text{ord}\,x|a,\text{ord}\,y|b\Rightarrow (\text{ord}\,x,\text{ord}\,y)=1$ so:
$$\text{ord}(x,y)=\text{ord}\,x\cdot\text{ord}\, y<a\,\text{ord}\, y\leq ab$$
Hence $(x,y)$ cannot possibly generate $C_{ab}$.
Proposition: $\sum_{d|n}\phi (d)=n$ 
Proof: By induction. $n=1:\;\sum_{d|1}\phi (d)=\phi(1)=1$. Assume $\sum_{d|k}\phi (d)=k$ holds for $n< k$. If $k=p^n,\;p$ prime, then: $$\begin{aligned}\sum_{d|k}\phi(d)&=\sum_{j=0}^n \phi(p^j)=1+\sum_{j=1}^n (p^j-p^{j-1})=p^n\end{aligned}$$ and we are done. Otherwise write $k=p^mq$ for some prime $p$, where $m$ is the highest power of $p$ dividing $k$ and $q>1$. Then $q<k$ and by our lemma: $$k=p^m q=\left(\sum_{d|p^m}\phi(d)\right)\left(\sum_{d|q}\phi(d)\right)=\sum_{j=0}^m\phi(p^j)\sum_{d|q} \phi(d)=\sum_{j=0}^m\sum_{d|q}\phi(p^j d)$$
Since each divisor of $k$ can be written $p^jd$ for some $d|q$ and $0\leq j\leq m$ this sum is over all divisors of $k$ (indeed, none are counted more than once). Hence $\sum_{d|k}\phi(d)=k$ and we are done.

I know there are other proofs of this, and that most are simpler, but this is what came most naturally to me when I set the problem to myself as an exercise. Can I just check that it works and that the argument is sufficiently clear? Are there any improvements I can make?
Edit: I realise now that it is a little odd to use a group theoretic argument for the lemma but not for the proposition $-$ this is because my initial attempt at proving the lemma (purely number theoretic) was flawed: after this was pointed out by benh, the current proof is what came to mind. I will leave the second part as it was.

REPLY [2 votes]: Your proof of the Proposition looks good.
For the proof of your Lemma you seem to have the right idea. But if one is nitpicky the proof is incomplete: The reason is, that you have shown the assertion only for prime powers. However the step from products of two prime powers two products of more prime powers is not an induction step, because you can't just plug in another number instead of the prime power. So just saying "the claim follows inductively" is arguably not enough.
Instead you could give a general formula for $\phi(n)$ and use it for a very short proof or you give an argument as you have done in the last part of your proof of the Proposition.