I want to prove that for $ -1\lt q_1 \le q_2 \lt 1 $ then the following holds:
$$\sum_{n=0}^{\infty}(-1)^n \sum_{k=0}^n q_1^k \cdot q_2^{n-k} = \frac1{(1+q_1)(1+q_2)}$$
By the first inequality, $ -1\lt q_1 \le q_2 \lt 1 \Rightarrow \vert q_1 \vert \lt 1, \vert q_2 \vert \lt 1$ then $\sum_{n=0}^{\infty}(-q_1)^n = \frac1{1+q_1}$ and $\sum_{n=0}^{\infty}(-q_2)^n = \frac1{1+q_2}$
Now for the Cauchy product, by definition, $$\sum_{n=0}^{\infty}(-q_1)^n \sum_{n=0}^{\infty}(-q_2)^n = \sum_{n=0}^{\infty} \sum_{k=0}^n (-q_1)^k \cdot (-q_2)^{n-k} = \sum_{n=0}^\infty (-1)^n \sum_{k=0}^n (q_1)^k \cdot (q_2)^{n-k}$$
Now, by Merten's, since both series $\sum_{n=0}^{\infty}(-q_1)^n$ and $\sum_{n=0}^{\infty}(-q_2)^n$ are absolutely convergent to $\frac1{1+q_1}$ and $\frac1{1+q_2}$ respectively, we have that their Cauchy product will converge to $\frac1{1+q_1} \cdot \frac1{1+q_2}$
Ergo,
$$\sum_{n=0}^{\infty}(-1)^n \sum_{k=0}^n q_1^k \cdot q_2^{n-k} =\frac1{1+q_1} \cdot \frac1{1+q_2}= \frac1{(1+q_1)(1+q_2)}$$
Is there anything missing on my proof?
Two of your $q_1^n$ should instead be $q_1^k$. Also, $$\sum_{n=0}^{\infty} \sum_{k=0}^n (-q_1)^k \cdot (-q_2)^{n-k} = \sum_{n=0}^\infty (-1)^n \sum_{k=0}^n q_1^k \cdot q_2^{n-k}$$ because you have already factored out $(-1)^n$.