Is there a methodical way to compute Euler's Phi function

Question

Is there an algorithmic or methodical way to "factorise" the numbers in euler's phi function such that it becomes easily computable?

For example, $\phi(7000) = \phi(2^3 \cdot 5^3 \cdot 7)$

I'm finding it difficult to find this "factorised" version of 7000 in terms that are easily computable in the phi function.

Also, since computing $\phi$ of a prime number is easy, I thought one method would be to factorise 7000 down to multiples of ONLY prime numbers, ie:

$7000 = 7 \times 5 \times 5 \times 5 \times 2 \times 2 \times 2 $
$\phi(7000) = \phi(7) \times \phi(5) \times \phi(5) \times \phi(5) \times \phi(2) \times \phi(2) \times \phi(2)$
$\phi(7000) = 6 \times 4 \times 4 \times 4 \times 1 \times 1 \times 1 $
(Since the $\phi$ of a prime number is just that number $- 1$)

But the answer is not correct. I get $384$. Supposed to be $2400$. Why does this not work?

Answer 1

The problem is that $\phi(a \cdot b) = \phi(a) \cdot \phi(b)$ works if $a$ and $b$ are relatively prime, but it doesn't work in general.

Thus, it is true that $\phi(7000) = \phi(2^3) \cdot \phi(5^3) \cdot \phi(7)$. However, we cannot break $\phi(2^3)$ down to $\phi(2)^3$. (You can check that $\phi(2^3) = 4$ and $\phi(2)^3 = 1$.)

Thus, the problem reduces to figuring out how to evaluate $\phi(p^k)$ where $p$ is a prime. There is a nice formula for this, which isn't very tricky to find. Try some examples. Do you see a pattern?

Answer 2

Are you aware if that $ \phi(m)=m \prod_{i=1}^k(1-\frac{1}{p_i})$ where $p_i$ is a prime in the canonical representation of m?

Edit:I did not see J.M.'s comment.