Let $f:\mathbb{N}\rightarrow\mathbb{N}$ be a strictly increasing function such that $$f(2)=2$$ and $$f(mn)=f(m)f(n)$$ for all positive integers $m,n$ such that $$\gcd(m,n)=1$$ Prove that $f(n)=n$ for every positive integer $n$.
This is the original question which I have solved by induction, but a comment is given that $f:\mathbb{N}\rightarrow\mathbb{C}$ with $f(1)=1$ and $f(mn)=f(m)f(n)$ whenever $m$ and $n$ are coprime is called a multiplication function. We have the general result that for any increasing multiplicative function that is not constant is of the form $$f(n)=n^\alpha$$ for some $\alpha > 0$
My initial approach to the original problem was to use group theory; is that how this additional, more general, result is derived? I thought that the map is a homomorphism and this preserves order of elements, but I wasn't sure how to go further.
Can anyone give a proof (to either the original problem or the more general result) in terms of this method? Or provide an alternate method?
