Representing the $q$-binomial coefficient as a polynomial with coefficients in $\mathbb{Q}(q)$? Trying a bit of combinatorics this winter break, and I don't understand a certain claim.

The claim is that for each $k$ there is a unique polynomial $P_k(x)$ of degree $k$ whose coefficients are in $\mathbb{Q}(q)$, the field of rational functions, such that $P_k(q^n)=\binom{n}{k}_q$ for all $n$.

Here $\binom{n}{k}_q$ is the $q$-binomial coefficient. I guess what is mostly troubling me is that $P_k(q^n)$ is a polynomial in $q^n$. I'm sure it's obvious, but why is the claim true? Thanks.

The $q$-binomial coefficient satisfies the recurrence
$$
  \binom{n}{k}_q = q^k \binom{n-1}{k}_q + \binom{n-1}{k-1}_q,
$$
which follows easily from the definition. We can assume inductively that each term on the right is a polynomial and therefore the LHS is a polynomial.
Edit: Unfortunately this does not seem to yield the uniqueness required in the question.
