I want to show the identity $\bigl[\!\begin{smallmatrix} n \\ k \end{smallmatrix}\!\bigr]_q=\bigl[\!\begin{smallmatrix} n-1 \\ k-1 \end{smallmatrix}\!\bigr]_q+q^k\bigl[\!\begin{smallmatrix} n-1 \\ k \end{smallmatrix}\!\bigr]_q$ for $0<k<n$ and the steps I followed are \begin{eqnarray*} \bigl[\!\begin{smallmatrix} n \\ k \end{smallmatrix}\!\bigr]_q-\bigl[\!\begin{smallmatrix} n-1 \\ k-1 \end{smallmatrix}\!\bigr]_q &=&\frac{(q^{n}-1)(q^{n-1}-1)\cdots(q^{n-k+1}-1)}{(q^{k}-1)(q^{k-1}-1)\cdots(q-1)}-\bigl[\!\begin{smallmatrix} n-1 \\ k-1 \end{smallmatrix}\!\bigr]_q\\ &=&\left(\frac{q^{n}-1}{q^{k}-1}-1\right)\bigl[\!\begin{smallmatrix} n-1 \\ k-1 \end{smallmatrix}\!\bigr]_q\\ &=&q^k\frac{q^{n-k}-1}{q^{k}-1}\bigl[\!\begin{smallmatrix} n-1 \\ k-1 \end{smallmatrix}\!\bigr]_q\\ &=&q^k\frac{(q^{n-k}-1)(q^{n-1}-1)\cdots(q^{n-k+1}-1)}{(q^{k}-1)(q^{k-1}-1)\cdots(q-1)}\\ &=&q^k\bigl[\!\begin{smallmatrix} n-1 \\ k \end{smallmatrix}\!\bigr]_q \end{eqnarray*}
Assuming that the rest of the derivation is correct, the only part I was not able to show is the last equation (which should hold) i.e. $$\frac{(q^{n-k}-1)(q^{n-1}-1)\cdots(q^{n-k+1}-1)}{(q^{k}-1)(q^{k-1}-1)\cdots(q-1)}=\bigl[\!\begin{smallmatrix} n-1 \\ k \end{smallmatrix}\!\bigr]_q$$ even though I am not sure whether this is the correct way to go about this.
Here's your numerator:
$$ \color{Red}{(q^{n-k}-1)}(q^{n-1}-1)\cdots(q^{n-k+1}-1)$$
Move the first term to the end of the product.