$f(1/n)=1/n$ implies $f(z)=z$

Question

If $f:\mathbb{C}\to\mathbb{C}$ is entire and satisfies $f(1/n)=1/n$ then $f(z)=z$.

I am trying to prove this. What I have so far: by continuity $f(0)=0$. Since $f(1)=1$, it feels tempting to show that $|f(z)|\leq 1$ on the unit disk. Then, by Schwarz' lemma, we would get $f(z)=az$ on the disk, and hence $f(z)=z$ since $f$ entire. My question is:

Does the condition indeed imply $f(z)\leq 1$, and if so, why?

Answer 1

Just so this does not go unanswered:

The identity theorem is the following:

If $f$ and $g$ are holomorphic on a domain $\Omega$, and $f = g$ on some subset $S \subset \Omega$, and $S$ has a limit point in $\Omega$, then $f = g$ on all of $\Omega$

The proof can be found on wikipedia:

https://en.wikipedia.org/wiki/Identity_theorem

The question in the OP is immediately answered by this theorem.