Deduce that ${{n} \brack {k}}_{q}$ is a symmetric polynomial of $q$, that is, if \begin{equation*} {{n}\brack {k}}_{q} = a_0 + a_1q + a_2q^2 + \ldots + a_Nq^N \end{equation*} with $a_N \neq 0$, then $a_i = a_{N-i}$ for all $i$.
I'm having trouble proving this. Is this simply by symmetry of the Gaussian binomial coefficient (${n \brack k} = {n \brack n-k}$)?
Edit: I found a much more elementary proof of this fact, the old proof I had is below the line.
A polynomial $p$ of degree $d$ is symmetric iff $p(x)=x^dp(x^{-1})$. The Gaussian binomial coefficient $\binom{n}{k}_q$ has degree $k(n-k)$, so the following proves it is symmetric: $$ \begin{align} q^{k(n-k)}\binom{n}{k}_{q^{-1}} &=q^{k(n-k)}\frac{(q^{-n}-1)(q^{-n+1}-1)\cdots(q^{-1}-1)}{(q^{-k}-1)\cdots(q^{-1}-1)(q^{-(n-k)}-1)\cdots (q^{-1}-1)}\\ &=\left(\frac{q^{n(n+1)/2}}{q^{k(k+1)/2}\cdot q^{(n-k)(n-k+1)/2}}\right)\cdot\frac{(q^{-n}-1)(q^{-n+1}-1)\cdots(q^{-1}-1)}{(q^{-k}-1)\cdots(q^{-1}-1)(q^{-(n-k)}-1)\cdots (q^{-1}-1)}\\ &=\frac{(1-q^{n})(1-q^{n-1})\cdots(1-q)}{(1-q^k)\cdots(1-q)(1-q^{n-k})\cdots (1-q)}=\binom{n}{k}_q \end{align} $$
You may recall that $\binom{n}{k}$ counts the number of lattice paths from $(0,0)$ to $(n-k,k)$ where every step is either up or right. This is because such a path can expressed as a string of $n$ letters, where $k$ of them are U for "up" and $n-k$ are R for "right."
The coefficients of the polynomial $\binom{n}{k}_q$ also have a meaning related to lattice paths. Specifically:
This is discussed in the wikipedia article.
If you can prove the above assertion, the symmetry of $\binom{n}{k}_q$ follows, since rotating a lattice path $180^\circ$ is a bijection from lattice paths with an area of $m$ to lattice paths with an area of $k(n-k)-m$.
You can prove this combinatorial interpretation by using the $q$-Pascal's identity $$ \binom{n}{k}_q=q^k\binom{n-1}{k}_q+\binom{n-1}{k-1}_q $$ and then showing that the number of lattice paths from $(0,0)$ to $(k,n-k)$ with an area of $m$ obeys a similar recurrnce.