TITLE: Unimodality of q-binomial coefficients
QUESTION [7 upvotes]: The q-Pochhammer symbol $[n]_q!$ is defined as
$$[n]_q! = \frac{(1-q^n)(1-q^{n-1})\cdots(1-q)}{(1-q)^n} = (1+q) (1+q+q^2) \cdots (1+q+\cdots+q^{n-1})$$
It can be easily shown that $[n]_q!$ (function of indeterminate $q$) is symmetric and unimodal, since it is the product of symmetric and unimodal sequences.
The q-binomial coefficient is defined similarly as, 
$${n \choose k}_q = \frac{[n]_q!}{[k]_q![{n-k}]_q!}$$
It can be easily seen that ${n \choose k}_q$ is a polynomial in $q$ (and not a rational polynomial in $q$) from the recurrence relation 
$${n \choose k}_q  = q^k {{n-1} \choose k}_q  + {{n-1} \choose {k-1}}_q $$
Can somebody show me an easy way to prove that ${n \choose k}_q$ is also symmetric and unimodal?
Edit: A polynomial in indeterminate $x$, F(x) = $\sum_{i=0}^n a_i x^i$ is defined to be symmetric when the sequence $\{a_i\}_{i=0}^n$ is unimodal and symmetric. The sequence $\{a_i\}_{i=0}^n$ is defined to be unimodal and symmetric if -
$$a_0 \leq a_1 \leq .... \leq a_{\lfloor{n/2}\rfloor}$$
and 
$$a_k = a_{n-k}$$
Now for ${n \choose k}_q = F(q)$, we have $F(q) = q^{{(nk-k^2)}} F(1/q)$ therefore, $F(q)$ must be symmetric (in the above sense). How do I show it to be unimodal?

REPLY [3 votes]: Symmetry is immediate from definition of $\binom{n}k_q$, since it’s symmetric in $k$ and $n-k$.
$$\begin{align*}
\binom{n}k_q&=\frac{[n]_q!}{[k]_q![n-k]_q!}\\\\
&=\frac{(1-q)^{-n}\prod_{j=1}^n\left(1-q^j\right)}{\left((1-q)^{-k}\prod_{j=1}^k\left(1-q^j\right)\right)\left((1-q)^{-(n-k)}\prod_{j=1}^{n-k}\left(1-q^j\right)\right)}\\\\
&=\frac{\prod_{j=1}^n\left(1-q^j\right)}{\left(\prod_{j=1}^k\left(1-q^j\right)\right)\left(\prod_{j=1}^{n-k}\left(1-q^j\right)\right)}\\\\
&=\frac{\prod_{j=n-k+1}^n\left(1-q^j\right)}{\prod_{j=1}^k\left(1-q^j\right)}\;,
\end{align*}$$
so
$$\binom{n}{k+1}_q=\binom{n}k_q\cdot\frac{q^{n-k}-1}{q^{k+1}-1}\;,$$
and $$\binom{n}{k+1}_q\ge\binom{n}k_q$$ iff $n-k\ge k+1$, i.e., iff $k\le\frac{n-1}2$, which is precisely the desired unimodality.