Here is a very elementary number theory proof using strong induction. Please mark/grade.

Prove that $$\sum_{d|n}\phi(d)=n$$where $\phi$ is the Euler's phi function, $n,d\in\mathbb{N}$

First, when n=1,$$\sum_{d|1}\phi(d)=\phi(1)=1$$
Second, assume $$\sum_{d|k}\phi(d)=k\;\forall k\lt n,\;k\in\mathbb{N}$$
and let $n=p^kq, \;p\nmid q, p$ is prime,

Then $$\sum_{d|n}\phi(d)$$
$$=\sum_{0\le j\le k,\;e|q}\phi(p^je)$$
$$=\sum_{0\le j\le k,\;e|q}\phi(p^j)\phi(e)$$
$$=\sum_{0\le j\le k}\phi(p^j)\sum_{e|q}\phi(e)$$
$$=(\phi(1)+\sum_{1\le j\le k}(p^j-p^{j-1}))q$$
$$=(1+p^k-1)q$$
$$=p^kq$$
$$=n$$
By strong induction, $$\sum_{d|n}\phi(d)=n\;\forall n\in\mathbb{N}$$
Is my proof ok? Also, is there some more elegant proof? Thank you.