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{title2=$q^{m^2}\prod_{i=1}^m(q^{2i}-1)$}

Counting successive symplectic pairs gives
$$
|\operatorname{Sp}_{2m}(q)|=q^{m^2}\prod_{i=1}^m(q^{2i}-1).
$$
For each nonzero first vector there are $q^{2m-1}$ partners with pairing one, after which its nondegenerate plane splits off. The exact defining-characteristic part is $q^{m^2}$, realized by an upper symplectic unitriangular subgroup.


Finite symplectic group order

Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 3 / 3 / b / Solution

Finite symplectic group order ( $q^{m^{2}} \prod_{i = 1}^{m} (q^{2 i} - 1)$ )

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Finite symplectic group order (qm2∏i=1m​(q2i−1))

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Finite symplectic group order ( $q^{m^{2}} \prod_{i = 1}^{m} (q^{2 i} - 1)$ )

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