I wish to derive, for a proof, the Euler Totient Function starting from the case $n$ prime to $n$ non-prime.
Let $n$ prime, we know $\Phi(n)=n-1$. But what if now I assume $n=p_0^{a_0}p_1^{a_1}...p_r^{a_r}$ and I want to reach $\Phi(n)=n \prod_{p|n}^{}(1-1/p) $ from there? Like:
$\Phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})=p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1 \Rightarrow p_0^{a_0}p_1^{a_1}...p_r^{a_r}(1-1/(p_0^{a_0}p_1^{a_1}...p_r^{a_r}))=?$
Is this kind of approach possible for this formula?
To be precise, your "kind" of derivation is not possible. This is because $\phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})\neq p_0^{a_0}p_1^{a_1}...p_r^{a_r}-1$.
Think about what the totient function entails: the number of integers less than the input that are coprime with it. Only with prime inputs is this one less than the input.
The totient function, however, does retain two properties, namely as reuns mentioned, that $\phi(p^k)=p^k-p^{k-1}$ and $\phi(n)\phi(m)=\phi(mn)$ when $m$ and $n$ are coprime. The proof of these is easily verifiable given a google search.
Thus, we can have a "direct" approach. Since powers of primes are always coprime to one another, \begin{align*} \phi(n)&=\phi(p_0^{a_0}p_1^{a_1}...p_r^{a_r})\\ &=\phi(p_0^{a_0})\phi(p_1^{a_1})...\phi(p_r^{a_r})\\ &=(p_0^{a_0}-p_0^{a_0-1})(p_1^{a_1}-p_1^{a_1-1})...(p_r^{a_r}-p_r^{a_r-1})\\ &=p_0^{a_0}(1-1/p_0)p_1^{a_1}(1-1/p_1)...p_r^{a_r}(1-1/p_r)\\ &=n\displaystyle\prod_{p|n}(1-1/p). \end{align*}