= Solution
Assume $0\le k\le n$ with $n,k$ integers, as required for the quantum binomial expression. Let $L_{n-1}$ be the $n$-dimensional irreducible <sl2 Lie algebra> representation, with one-dimensional <weight spaces> of <weights> $n-1-2j$, $0\le j<n$. The <exterior-power Lie algebra representation> on $\Lambda^kL_{n-1}$ has character
$$
C_{n,k}(q)=\sum_{0\le j_1<\cdots<j_k<n}
q^{k(n-1)-2(j_1+\cdots+j_k)}.
$$
This is already a <Laurent polynomial> with nonnegative integer coefficients.
To identify it with the requested quotient, introduce $t=q^{-2}$ and the <Gaussian binomial coefficient>
$$
G_{n,k}(t)=\sum_{0\le j_1<\cdots<j_k<n}
t^{j_1+\cdots+j_k-k(k-1)/2}.
$$
Splitting the subsets according to whether $n-1$ is included proves
$$
G_{n,k}=G_{n-1,k}+t^{n-k}G_{n-1,k-1},\qquad
G_{n,0}=G_{n,n}=1.
$$
The product $\prod_{j=1}^k(1-t^{n-k+j})/(1-t^j)$ satisfies the same recurrence: after factoring common numerator and denominator terms, the needed identity is $(1-t^{n-k})+t^{n-k}(1-t^k)=1-t^n$. Induction on $n$ therefore proves the product identity, including its polynomiality.
Since the <quantum integer> is $[a]_q=q^{a-1}(1-q^{-2a})/(1-q^{-2})$, taking the quotient of the products gives
$$
\boxed{\left[\begin{matrix}n\\k\end{matrix}\right]_q
=q^{k(n-k)}G_{n,k}(q^{-2})
=C_{n,k}(q)=\chi_{\Lambda^kL_{n-1}}(q).}
$$
Thus this is the <symmetric quantum binomial coefficient>. All its <weights> have parity $k(n-k)$, because the exponent in the exterior-power sum differs from $k(n-k)$ by an even integer. Applying part i to this finite-dimensional representation proves \b[symmetric unimodality in steps of two]. Equivalently, the ordinary <polynomial> $G_{n,k}(t)$ has a unimodal coefficient sequence at consecutive powers of $t$.
Again, the full integer-exponent coefficient sequence in $q$, including the absent parity, need not be unimodal: for $n=2,k=1$ the answer is $q+q^{-1}$. This counterexample makes precise the convention under which the claimed quantum unimodality holds. The cases $k=0$ or $k=n$ give the constant one.
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