eulers product form

Question

I have to prove the following: $\varphi(n)=n\prod\limits_{p\mid n}(1-\frac{1}{p})$.

To prove this I have to use the following theorem: Let n,r,a$_{1}$,...,a$_{r}$ are pairwise coprime, the number of natural numbers < n divisible by none of a$_{i}$ is equal to

n-$\sum\limits_{i=1}^{r}\lfloor\frac{n}{a_{i}}\rfloor$+$\sum\limits_{1\leq i<j\leq r}\lfloor \frac{n}{a_{i}a_{j}}\rfloor\pm\cdot\cdot\cdot+(-1)^{r}\lfloor\frac{n}{a_{1},...,a_{r}}\rfloor$.

I choose a$_{1}$,...,a$_{r}$ as the enumeration of prime divisors of n. Then for i$\neq$j is $\lfloor\frac{n}{a_{i}a_{j}}\rfloor = \frac{n}{a_{i}a_{j}}$.

Then I think I have to rearrange this

n-$\sum\limits_{i=1}^{r}\lfloor\frac{n}{a_{i}}\rfloor$+$\sum\limits_{1\leq i<j\leq r}\lfloor \frac{n}{a_{i}a_{j}}\rfloor\pm\cdot\cdot\cdot+(-1)^{r}\lfloor\frac{n}{a_{1},...,a_{r}}\rfloor$ to this n$\prod\limits_{p\mid n}(1-\frac{1}{p})$, but I can't.

Can you help me rearrange this?

Thanks in advance

Does this answer your question? Proving that $\varphi(n)=n\prod (1-1/p)$ without using multiplicativity. The answer by Indrayudh Roy uses inclusion-exclusion - very nice. I think this is what you need. — Dietrich Burde, Feb 23 '22 at 13:24
If the $a_i$s are prime divisors of $n$ then $n$ and $a_i$ are not coprime and the $\lfloor \frac{n}{a_{i}a_{j}}\rfloor$ etc. do not need to be rounded down — Henry, Feb 23 '22 at 13:41

eulers product form

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