Very elementary proof of that Euler's totient function is multiplicative Well, I know two or three proofs of this fact
$$\gcd(m,n)=1\implies \varphi(mn)=\varphi(m)\varphi(n)$$
where $\varphi$ is the totient function.
My problem is this: I'd like to explain this to some gifted children. The children are gifted enough to understand some basic facts, like the reason why $\varphi(n)$ is even ($k$ and $n$ are coprime $\iff$ $n-k$ and $k$ are, so make pairs) and they have a very solid intuitive idea of the Fundamental Theorem of Arithmetic (they know, for example, that if $a$ and $b$ are coprime, and $ab$ is a square, then both $a$ and $b$ are squares).
But I have tried to taught them why the totient function is multiplicative, but I (and they)  understand that $\varphi$ is multiplicative but not an intuitive reason.
How would you explain it?

Here's an intuitive but informal argument. For any fixed $n$, one has that $\frac{\varphi(n)}n$ is the probability that any arbitrarily chosen integer (uniformly in a very large interval) is relatively prime to$~n$. The multiplicativity of$~\varphi$ says that if $m,n$ have no common prime factors, then the events of being relatively prime to $m$ and to$~n$ are independent. It is easy to see that being divisible by one prime number is independent of being divisible by another prime number, and this can be generalised to the mutual independence of divisibility by any set of prime numbers. The multiplicativity of$~\varphi$ is now clear, because the sets of prime factors of $m,n$ are disjoint.
Of course making this into a formal argument requires proving the Chinese remainder theorem.
