Extending the power of Totient function

Up to this point we know that $\phi(p)$ for $p$ prime is $\phi(p) = p - 1$, and that the multiplicative property holds between primes.
Now we want to extend this to non-primes. We know that every non-prime is a product of powers of primes, then the first thing to do is know how the Totient function behaves for powers of primes.
Let's take $p^{k}$. We need to calc. n. of coprimes.
To solve this in the easiest way possible we can remove those we are sure are not coprimes. Every number multiplied by $p$ which produces $p^{k}$ is not coprime to $p^{k}$, while every other number from 1 to $p^{k}$ is coprime instead, because $p$ is prime and is only mult. by itself.
Then: $n * p = p^{k}$ -> $n = p^{k} / p$ -> $n = p^{k - 1}$
Then: $p^{k-1}$ are not coprimes -> $p^{k} - p^{k-1}$ are coprimes -> $\phi(p^{k}) = p^{k} - p^{k-1}$

Now the ¿hard? stuff

We know that multiplicativity holds for primes, but here we have $\phi(p^{k}) = p^{k} - p^{k-1}$ where $p^{k}$ is not a prime.
If we manage to demonstrate that multiplicativity holds for non-primes (powers of primes actually) we can multiply $\phi(p^{k})$ for every $p_{i}, k_{i}$ and derive the property for non-primes. This will mean being able to calculate Totient for every integer and being sure the calculation is correct.

Example: $\phi(2^{20}5^{15})$

Let's look at the new scheme for this example. Making this has helped me a lot in the process.

Here becomes clear that removing co-factors will be harder. The previous reasoning scheme holds but now we will have to remove all the columns and lines which are mults of $2$ and $5$, and for every line and column removed we will need to remove $1$ element considered twice and so intersections will be a huge? problem. So we will need to remove for ex: $5^{15}(2)$ column, which is composed of $5^{15}$ elements which share the $2$ factor, the same goes for $2^{20}(5)$ line, here there will be $2^{20}$ elements to remove because they share the $5$ factor.
We know that we will have to remove $2^{20} / 2$ columns + $5^{15} / 5$ lines in total, this means: $(2^{20} / 2) * 5^{15} + (5^{15} / 5) * 2^{20}$ and since we will need to sub $1$ for every intersection, we will need to sub $5^{15} / 5 * 2^{20} / 2$. Let's write this whole formula:

$\phi(2^{20}5^{15}) = 5^{15} × 2^{20} − ((2^{19} × 5^{15}) + (5^{14} × 2^{20}) − 5^{14} × 2^{19})$

Formalization

$\phi(x^{k}y^{z}) = x^{k}y^{z} - x^{k-1}y^{z} - x^{k}y^{z-1} + x^{k-1}y^{z-1}$
$x^{k}y^{z} - x^{k-1}y^{z} - x^{k}y^{z-1} + x^{k-1}y^{z-1} = (x^{k} - x^{k-1})(y^{z} - y^{z-1})$

where

$\phi(x^{k}) = x^{k} - x^{k-1}$
$\phi(y^{z}) = y^{z} - y^{z-1}$

Refinition of the scheme ensuring formalization is unbreakable

The scheme above is a little bit messy, and looks like it's not following a structured reasoning, this time is going to be harder to make it structured and perform swaps and preserve formalization. Structured scheme:

Looking at this scheme it's easy to see that $2^{19} × 5^{15}$ columns are removable. It's not the same for $5^{14} × 2^{20}$. To remove the latter we will need swaps. Of course this time it won't be possible to show the complete picture and the reasoning will need to abstract the image representation.
We already know that there are $5^{14}$ elements among lines to be removed. The multiplication by $2^{20}$ is not really trivial too: all these cofactors are mutiplied by all numbers from $1$ to $2^{20}$ and the products are obviously elements which share $5$ co-factor and then they all need to be removed. Then we can imagine swaps between lines which preserve the formalization. $5^{14} × 2^{19}$ is the last term we need to match the formalization and is a consequence of this whole reasoning (intersections).