= Solution
Let $q$ be a prime power and let $V=\mathbb F_q^{2m}$ carry a <nondegenerate> <alternating bilinear form> $B$. The <symplectic group over a finite field> is
$$
Sp(V,B)=\{g\in GL(V):B(gv,gw)=B(v,w)\text{ for all }v,w\}.
$$
In a <symplectic basis> its <matrix> description is $g^{\mathsf T}Jg=J$, where $J=\left(\begin{smallmatrix}0&I_m\\-I_m&0\end{smallmatrix}\right)$. The definition is valid in characteristic two: alternating means $B(v,v)=0$, and then the form is also symmetric.
Count ordered <symplectic bases> $(e_1,f_1,\ldots,e_m,f_m)$ with $B(e_i,f_j)=\delta_{ij}$ and all $e$-$e$ and $f$-$f$ pairings zero. There are $q^{2m}-1$ choices for $e_1\ne0$. Nondegeneracy makes $v\mapsto B(e_1,v)$ a nonzero <linear functional>, so there are $q^{2m-1}$ choices for $f_1$ satisfying $B(e_1,f_1)=1$.
The span $P=\langle e_1,f_1\rangle$ is <nondegenerate>. Thus $V=P\oplus P^\perp$, and its <bilinear orthogonal complement> is <nondegenerate> of dimension $2m-2$. For completeness, a <vector> in the <radical of a bilinear form> of $P^\perp$ is orthogonal both to $P$ and to $P^\perp$, hence to all $V$ and therefore zero. Recursing constructs and counts every <symplectic basis>. If $b_m$ is their number, $b_m=(q^{2m}-1)q^{2m-1}b_{m-1}$ with $b_0=1$.
The <symplectic group over a finite field> acts freely and transitively on <symplectic bases>: there is a unique <linear map> sending one ordered <basis> to another, and its equality of <basis> pairings ensures it preserves $B$. Its order is consequently $b_m$, giving
$$
\boxed{|Sp(2m,q)|=\prod_{i=1}^mq^{2i-1}(q^{2i}-1)=q^{m^2}\prod_{i=1}^m(q^{2i}-1).}
$$
The exponent is $1+3+\cdots+(2m-1)=m^2$. The counting is valid for every prime power $q$, including $q=2$.
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