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Oct 28 at 1:30 comment added Vladimir Dotsenko Dear Neil, I have been telling a lot of people about your argument, and I am informed that one of those conversations was fruitful, and there is an arXiv preprint that proves a much more intricate thing this way: arxiv.org/abs/2409.05605 !
Dec 15, 2022 at 9:42 history edited Neil Strickland CC BY-SA 4.0
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Dec 15, 2022 at 9:34 comment added Neil Strickland OK, I missed the point that you have a product of terms $1-q^i$ rather than $q^i-1$; I will edit my answer.
Dec 15, 2022 at 9:30 vote accept Vladimir Dotsenko
Dec 15, 2022 at 9:30 comment added Vladimir Dotsenko Thanks Neil! This is lovely. Though with my conventions the identity is correct (since my $(q;q)_k$ evaluates to $1$ at $q=0$): the cardinality of $GL_k$ is $(-1)^kq^{\binom{k}2}(q;q)_k$ ;)
Dec 15, 2022 at 9:26 history answered Neil Strickland CC BY-SA 4.0