For whatever reason, I was sifting through an elementary number theory book and stumbled upon a proof for the theorem that $\phi$ is a multiplicative function. I did not like the proof that used an array and/or the Chinese Remainder Theorem, so I set out to prove this in the following way:
Lemma $1:$ \begin{align} \phi (p^k) = p^{k}-p^{k-1} \end{align}
Lemma $2:$ \begin{align} \phi (p^kq^j) = \phi(p^k) \phi(q^j)\end{align} which is trivial from Lemma $1$.
It seems that if I can get a way to characterize any $n$, and a way to induct through the prime powers, the result follows in a simple way through rearranging various prime powers in the prime factorization of $n$ or evaluated at those prime powers.
Lemma $3:$ \begin{align}\phi (p_1^k\cdot\cdot\cdot p_n^{k_n})= \phi (p_1^k)\cdot\cdot\cdot \phi(p_n^{k_n}) \implies \phi(p_1^k\cdot\cdot\cdot p_{n+1}^{k_{n+1}})= \phi (p_1^k)\cdot\cdot\cdot \phi(p_{n+1}^{k_{n+1}})\end{align} where, if another prime power is appended, the expression can still be split in the same way. It does not seem to work and makes me think, upon seeing the $n=2$ case, that the whole operation can be simplified to an inclusion-exclusion problem on the number of smaller relatively prime numbers to some product of $n$ prime powers.
If there is some way to use an induction argument to finish this, please explain how. If not, please help by providing a combinatorial expression for the general $n$ case of the first equality in Lemma $3$.It escapes me whether or not such an expression can be written in a trivial, compact manner, but it appears that this can be done combinatorially and then verified relatively simply using Lemma $1$.
