${n \brack k} = \sum_{m=0}^{k}2^{(k-m)^2}{n-k \brack k-m}{k \brack m}$ $${n \brack k} = \sum_{m=0}^{k}2^{(k-m)^2}{n-k \brack k-m}{k \brack m}$$ I need hint to prove this. ${n \brack k}$ is the number of $k$ dimension...--prophetes.ai

${n \brack k} = \sum_{m=0}^{k}2^{(k-m)^2}{n-k \brack k-m}{k \brack m}$ $${n \brack k} = \sum_{m=0}^{k}2^{(k-m)^2}{n-k \brack k-m}{k \brack m}$$ I need hint to prove this. ${n \brack k}$ is the number of $k$ dimension subspaces of $n$ dimension space over field $F_2$. I am thinking about using induction and Pascal identities for the Gaussian binomial coefficients but cant figure out how.

It turned out this is special case of q-Vandermonde identity. Proof can be found here: <