Gaussian polynomial identities I'd appreciate any hints for showing that these identities are true for Gaussian polynomials. I've tried to approach the problem using basic algebra but it gets messy very quickly and I've gotten stuck multiple times now. Thanks in advance.


*

*$
{n \brack k}=\frac{1-q^n}{1-q^{n-k}}{n-1 \brack k}$

*$
{n \brack k} = {n-1 \brack k} +q^{n-k}{n-1 \brack k-1}
$


Edit: I'd like to add that I cannot use the $q$-binomial identity.
 A: Here we use the following settings:
\begin{align*}
[k]_q:=\frac{1-q^k}{1-q}\qquad\qquad [k]_q!:=\prod_{j=1}^{k}[j]_q=\prod_{j=1}^{k}\frac{1-q^j}{1-q}
\end{align*}
The q-binomial coefficient $\begin{bmatrix}n\\k\end{bmatrix}_q$ is defined as
\begin{align*}
\begin{bmatrix}n\\k\end{bmatrix}_q:=\frac{[n]_q!}{[k]_q![n-k]_q!}
\end{align*}

Based upon these definititions we can show the following is valid
  \begin{align*}
\begin{bmatrix}n\\k\end{bmatrix}_q&=\frac{1-q^n}{1-q^{n-k}}\begin{bmatrix}n-1\\k\end{bmatrix}_q\tag{1}\\
\\
\begin{bmatrix}n\\k\end{bmatrix}_q&=\begin{bmatrix}n-1\\k\end{bmatrix}_q+q^{n-k}\begin{bmatrix}n-1\\k-1\end{bmatrix}_q\tag{2}
\end{align*}

$$ $$

We obtain
  \begin{align*}
\begin{bmatrix}n\\k\end{bmatrix}_q&=\frac{[n]_q!}{[k]_q![n-k]_q!}\\
&=\frac{[n]_q[n-1]_q!}{[k]_q![n-k]_q[n-k-1]_q!}\\
&=\frac{[n]_q}{[n-k]_q}\begin{bmatrix}n-1\\k\end{bmatrix}_q\\
&=\frac{1-q^n}{1-q}\frac{1-q}{1-q^{n-k}}\begin{bmatrix}n-1\\k\end{bmatrix}_q\\
&=\frac{1-q^n}{1-q^{n-k}}\begin{bmatrix}n-1\\k\end{bmatrix}_q
\end{align*}
  and (1) follows.

$$ $$

We obtain
  \begin{align*}
\begin{bmatrix}n-1\\k\end{bmatrix}_q&+q^{n-k}\begin{bmatrix}n-1\\k-1\end{bmatrix}_q\\
&=\frac{[n-1]_q!}{[k]_q![n-1-k]_q!}+q^{n-k}\frac{[n-1]_q!}{[k-1]_q![n-k]_q!}\\
&=\frac{[n]_q!}{[k]_q![n-k]_q!}\left(\frac{[n-k]_q}{[n]_q}+q^{n-k}\frac{[k]_q}{[n]_q}\right)\\
&=\begin{bmatrix}n\\k\end{bmatrix}_q\left(\frac{1-q^{n-k}}{1-q}\cdot\frac{1-q}{1-q^n}
+q^{n-k}\frac{1-q^k}{1-q}\frac{1-q}{1-q^n}\right)\\
&=\begin{bmatrix}n\\k\end{bmatrix}_q\left(\frac{1-q^{n-k}}{1-q^n}
+q^{n-k}\frac{1-q^k}{1-q^n}\right)\\
&=\begin{bmatrix}n\\k\end{bmatrix}_q
\end{align*}
  and (2) follows.

Note the similarity of algebraic manipulations when showing the validity of
\begin{align*}
\binom{n}{k}=\frac{n}{n-k}\binom{n-1}{k}\qquad\text{ and }\qquad\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}
\end{align*}
