Induction shows that an equality holds for all values of $n$. It doesn't show that this is the only equality or formula for the expression that may hold true, correct? For example, say a question asks to find an explicit formula for a functional equation given by $f(n) = f(f(n-1))+f(n-f(n-1))$ for $n > 1$. You can prove by induction that $f(n) = n$ is indeed an explicit expression for $f(n)$, but it may not be the only one? If so, what steps must be taken to prove that it is the only one?
Edit: Sorry, I meant to point out that $f(1) = 1$
Induction proves . . . whatever it proves. For instance:
We might use an induction argument to show that some $f$ is a solution to a functional equation $E$. In the absence of any other arguments, this won't say that $f$ is the unique solution.
On the other hand, we might use induction to show that $f$ is the unique solution to a functional equation $E$. For a silly example, if $E$ is $$f(n+1)=f(n)+1,$$ we can prove by induction that every solution $f$ to $E$ is of the form $f(x)=x+c$ for some fixed $c$. (Specifically, we would argue: let $f$ satisfy $E$, and set $c=f(0)$. Then by induction on $n$, we can show $f(n)=n+c$. So every solution to $E$ has this form.)
EDIT: in the specific case you're interested in, we can indeed use induction to prove that $f(n)=n$ is the only solution. There's a few ways to phrase it, but here's one which is general enough to be useful in other contexts:
We have an initial value, $f(1)=1$. (I'm assuming your natural numbers start with 1, not 0.)
Informally: In the functional equation, each value of $f(n)$ (for $n>1$) is determined by the values of $f$ on numbers $<n$.
Formally: By induction, we can show that if $f$ and $g$ are two solutions to the functional equation and $f(1)=g(1)$, then $f(n)=g(n)$ for every $n$; that is, $f=g$. (This is a good exercise.)
So there is a unique solution.