Euler's totient $\phi$ function as a product of primes

Question

Many web pages say that Euler's totient function $\phi(n)$ can be given as

$$\phi(n)=n \prod_{p|n} \biggl(1- \frac{1}{p} \biggr)$$

But $\phi(1)=1$, and no primes divide $1$. Surely this gives

$$\phi(1)=\prod_{p|1} \biggl(1- \frac{1}{p} \biggr)=0n=0$$

Is $\phi(1)$ a special case, or am I missing something?

Answer 1

There's no problem here.

$\phi(1)=1$ even according to the product definition;

there are no primes dividing $1$, so it's an empty product, which is $1$.

Answer 2

What is the sum of zero numbers? $0,$ of course. Most people have no trouble conceptually with this. Obviously $x + 0 = x$. So $0$ is the identity element of addition.

Now, what is the product of zero numbers? This one I need to go through the logic to convince myself of the right answer; I think most people do. The identity element is not $0$ because $x \times 0 = 0$, not $x$. But if we try $1$ instead, then bingo: $x \times 1 = x$.

Therefore, since no primes divide $1$, and if $\phi(n)$ really is a product of numbers of the form $$\frac{p - 1}{p}$$ with $p$ prime, it follows that $\phi(1) = 1$.

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There are good reasons why $1$ is not a prime number, and there are also stupid reasons. With Euler's totient function, we get a hint of one of the good reasons why $1$ is not a prime number: because in many fundamental ways, it behaves fundamentally differently from the prime numbers. Clearly $$1 - \frac{1}{1} = 1 - 1 = 0,$$ so then $\phi(n)$ would be $0$ for all $n$. Then $1$ being prime would require a special case for Euler's totient function to be of any use.

Lastly, I wanted to mention $\phi(0) = 0$. Since $0$ is divisible by all primes, even limiting to the positive primes, we'd have an infinite product on our hands. But then the whole thing is multiplied by $0$, so there is no need to even begin computing an infinite product...