= Solution
Assume $0\le k\le n$. Let $V=L_{n-1}$ be the $n$-dimensional irreducible <sl2 Lie algebra> representation. Choose a weight basis $v_0,\ldots,v_{n-1}$ with weights $2j-(n-1)$; this lists the usual irreducible weight string in increasing order. Its <exterior power> $\Lambda^kV$ has basis $v_{j_1}\wedge\cdots\wedge v_{j_k}$ for $0\le j_1<\cdots<j_k<n$, so its <formal character> is
$$
\operatorname{ch}\Lambda^kV
=\sum_{j_1<\cdots<j_k}q^{2(j_1+\cdots+j_k)-k(n-1)}
=q^{-N}\sum_{j_1<\cdots<j_k}(q^2)^{j_1+\cdots+j_k-k(k-1)/2},\quad N=k(n-k).
$$
The last sum is the <Gaussian binomial coefficient> $G_{n,k}(q^2)$. To verify the product identity, splitting the subsets according to whether they contain $n-1$ gives
$$
G_{n,k}(t)=G_{n-1,k}(t)+t^{n-k}G_{n-1,k-1}(t),\qquad G_{n,0}=G_{n,n}=1.
$$
The product $\prod_{j=1}^k(1-t^{n-k+j})/(1-t^j)$ satisfies the same recurrence, after putting both terms over their common denominator. Thus induction identifies it with the subset sum, proving it is a polynomial with nonnegative integer coefficients.
The <quantum integer> convention in the PDF has
$$
[m]_q=\frac{q^m-q^{-m}}{q-q^{-1}}=q^{1-m}\frac{1-q^{2m}}{1-q^2}.
$$
Multiplying these factors in the factorial quotient gives
$$
\boxed{\begin{bmatrix}n\\k\end{bmatrix}_q=q^{-N}G_{n,k}(q^2)=\operatorname{ch}\Lambda^kL_{n-1}.}
$$
This establishes the required representation-theoretic interpretation.
By the <Weyl complete reducibility theorem>, write $\Lambda^kV\cong\bigoplus_{m\ge0}r_mL_m$, where $r_m$ are nonnegative integers. The <classification of finite-dimensional sl2 representations> gives
$$
\operatorname{ch}L_m=q^m+q^{m-2}+\cdots+q^{-m}.
$$
Every exterior-power weight has parity $N$ and lies between $-N$ and $N$. Therefore only $m\le N$ with $m\equiv N\pmod2$ occur. For $j\ge0$ with that parity, the coefficient $c_j$ of $q^j$ is
$$
c_j=\sum_{\substack{m\ge j\\m\equiv N\ (2)}}r_m,\qquad c_j=c_{-j},\qquad c_j-c_{j+2}=r_j\ge0.
$$
The coefficients consequently increase toward the middle of the occupied weight string and decrease afterward. If $G_{n,k}(t)=\sum_{s=0}^N a_st^s$, then $a_s=c_{-N+2s}$. Hence
$$
\boxed{a_0\le a_1\le\cdots\le a_{\lfloor N/2\rfloor},\qquad a_s=a_{N-s}.}
$$
This is the <unimodality of Gaussian binomial coefficients>, proved through <parity unimodality of an sl2 character>. It includes $k=0,n$, when the character is one.
There is a normalization qualification in the printed wording. The symmetric quantum-binomial quotient is a <Laurent polynomial> in $q$, with exponents spaced by two; the unimodality just proved is for that occupied parity, or equivalently for the ordinary polynomial $G_{n,k}(t)$ after setting $t=q^2$ and removing the monomial shift. If one instead includes every integer exponent of $q$, the literal coefficient claim is false: $\begin{bmatrix}2\\1\end{bmatrix}_q=q+q^{-1}$ has coefficients $1,0,1$. \b[The representation-theoretic conclusion is unimodality on the parity string, and ordinary unimodality of the Gaussian polynomial.]
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