TITLE: Show that there exists no strictly increasing function $f:\mathbb{N}\rightarrow\mathbb{N}$ with $f(2)=3$...
QUESTION [7 upvotes]: Full exercise:

Show that there exists no strictly increasing function $f:\mathbb{N}\rightarrow\mathbb{N}$ with $f(2)=3$ which has the property that $f(mn)=f(m)f(n)$.

This is one of the first exercises in Putnam and Beyond, in the section dedicated to the proof by contradiction. 
I am quite familiar with the proof technique and feel comfortable with the mechanics of the problem but I find a nice trick here quite elusive. If anyone sees the elephant in the room a bit of subtle guidance would be much appreciated. I have the solution on hand but I would rather not look at it (where's the fun?). If you have any general suggestions that come to mind on how to handle proofs of this nature, especially those involving multiplicative homomorphisms between subsets of $\mathbb{R}$, I am all ears. Many thanks! 
P.S. I forgot to mention that I have tried using the fact that $f$ increasing implies $f(n+1)>f(n)$ for all $n\in\mathbb{N}$ in several ways to produce a contradiction of the $f(2)=3$ condition. Mainly I used the factorization of $n^2-1$ to get $f(n+1)f(n-1)<f(n)f(n)$ from the inequality $f(n^2)>f(n^2-1)$ but did not find anything very helpful in this approach.

REPLY [8 votes]: We have $f(4)=9,$ so $f(3)=4,5,6,7,8$.  We also have $f(8)=27$, so $f(9) \gt 27$ and $f(3) \gt 5$.  If $f(3)=8, f(27)=512$, but $f(32)=3^5=243$.  If $f(3)=7, f(81)=2401,$ but $f(128)=2187$.  If $f(3)=6, f(243)=7776,$ but $f(256)=6561$