I'm attempting to compute a repeated geometric sum
$$ \sum_{k_0=0}^m \alpha^{-k_0} \sum_{k_1=0}^{k_0} \alpha^{-k_1} \sum_{k_2=0}^{k_1} \alpha^{-k_2} \dots \sum_{k_q = 0}^{k_{q-1}} \alpha^{-k_q}. $$
If you start to explicitly compute the sums from right to left you see a pattern. It seems as the following is true
$$ \sum_{k=0}^{m} \alpha^{-nk} \prod_{l=1}^{n-1} \big( 1 - \alpha^{k + l} \big) = \frac{\alpha^{-nm}}{1 - \alpha^{n}} \prod_{l = 1}^{n} \big( 1 - \alpha^{m + l} \big) \quad \alpha \neq 1, \; \forall n \in \mathbb{N} \backslash \{0\} $$
Is this true? If so, how do one prove that is true?
I attempted to prove this using induction. The base case $n=1$ is true since
$$ \sum_{k = 0}^m \alpha^{-k} = \frac{\alpha^{-m} \big( 1 - \alpha^{m+1} \big)}{1 - \alpha}. $$
Assume that the formula holds for $n$ and let us show that it then must hold for $n+1$.
$$ \sum_{k=0}^{m} \alpha^{-(n+1)k} \prod_{l=1}^{n} \big( 1 - \alpha^{k + l} \big) = \sum_{k=0}^{m} \alpha^{-(n+1)k} \prod_{l=1}^{n-1} \big( 1 - \alpha^{k + l} \big) - \alpha^n \sum_{k=0}^{m} \alpha^{-nk} \prod_{l=1}^{n-1} \big( 1 - \alpha^{k + l} \big)\\ = \sum_{k=0}^{m} \alpha^{-(n+1)k} \prod_{l=1}^{n-1} \big( 1 - \alpha^{k + l} \big) - \alpha^n \frac{\alpha^{-nm}}{1 - \alpha^n} \prod_{l = 1}^n \big( 1 - \alpha^{m + l} \big) $$
However, from here I don't know how to proceed.