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If $n=p_1^{a_1}p_2^{a_2}...p_l^{a_l}$, then $\phi(n)=n(1-1/p_1)(1-1/p_2)...(1-1/p_l)$

Proof: Since $n=\sum_{d|n}\phi(n)$, Mobius inversion thm implies $$\phi(n)=\sum_{d|n}\mu(n){n\over d}=n-\sum_in/p_i+\sum_{i\lt j}{n\over p_ip_j}-...=n(1-1/p_1)(1-1/p_2)...(1-1/p_l)$$

I don't know how to convert all the summations into products. Any hint?

Thank you!

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If $f(n) = \sum_{d | n} g(d)$ is multiplicative then $g(n)=\sum_{d | n} \mu(d) f(n/d)$ is multiplicative.

Here $f(n) = n, g(n) =\phi(n)$. Multiplicative-ness means that $$g(n) = \prod_{p^k \| n} g(p^k) = \prod_{p^k \| n} (p^k - p^{k-1}) =\prod_{p^k \| n} p^k (1 - \frac{1}{p}) = n\prod_{p^k \| n} (1 - \frac{1}{p})$$ (where $p^k \| n$ means $p$ is prime, $p^k | n$ and $p^{k+1} \nmid n$)

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