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.

Existence. Recall $[\begin{smallmatrix} n \\ k \end{smallmatrix}]_q$ counts the $k$-dimensional subspaces of $\mathbb{F}_q^{\,n}$. A $k$-frame is an ordered linearly independent $k$-subset of $\mathbb{F}_q^{\,n}$. The number of $k$-frames may be counted in two ways: first, pick a vector, then a vector not in the span of the first, then a vector not in the span of the first two, and so on yielding $(q^n-1)(q^n-q)\cdots(q^n-q^{k-1})$. On the other hand, we may also count them by first picking a $k$-dimensional subspace, and then within that subspaces repeating the same process, yielding instead $[\begin{smallmatrix} n \\ k \end{smallmatrix}]_q(q^k-1)(q^k-q)\cdots(q^k-q^{k-1})$. Equating yields the formula 
$$ \begin{bmatrix} n \\ k \end{bmatrix}_q = \frac{\color{Red}{q^n}-1}{q^k-1}\frac{\color{Red}{q^n}-q}{q^k-q}\cdots\frac{\color{Red}{q^n}-q^{k-1}}{q^k-q^{k-1}}. $$
Evidently this is a polynomial in $q^n$ with coefficients in $\mathbb{Q}(q)$. By cancelling powers of $q$ in each of the fractions, then dividing top and bottom of each by $(q-1)$ in order to obtain $q$-analogs of integers, we may check this matches the textbook formula
$$ \frac{[n]_q[n-1]_q\cdots[n-(k-1)]_q}{[k]_q\,[k-1]_q\cdots\cdots\cdots\cdots[1]_q} = \frac{[n]_k!}{[k]_q!\,[n-k]_q!}. $$
Uniqueness. If $P,Q\in\mathbb{Q}(q)[T]$ and $P(q^n)=Q(q^n)$ for all $n$, then their difference $R(q^n)$ must be the zero element of $\mathbb{Q}(q)$ for all $n$. Without loss of generality, $R(T)\in\mathbb{Q}[q,T]$ after clearing denominators of coefficients of powers of $T$. Writing $R(T)=a_d(q)T^d+\cdots+a_1(q)T+q_0(q)$, we can find an $n$ large enough so that $\deg(a_d)+nd>\deg(a_k)+nk$ for $k<d$, so $\deg P(q^n)>0$, a contradiction, unless $d=0$ and so $R(T)=0$.
