 Solution

Write the alternating bilinear form as $B$. Alternation gives $B(v,w)=-B(w,v)$ by expanding $B(v+w,v+w)=0$. If $V\ne 0$, choose $v\ne 0$. Nondegeneracy supplies $w$ with $B(v,w)\ne 0$; rescale $w$ so that $B(v,w)=1$. The vectors are independent, and the restriction to $E=\mathrm{span}\{v,w\}$ has matrix $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, whose determinant is one.

Every $z\in V$ has a unique decomposition into a vector in $E$ and a vector orthogonal to $E$: subtract $B(z,w)v-B(z,v)w$ to obtain the latter. Thus $V=E\oplus E^{\perp }$. If a vector in $E^{\perp }$ annihilates $E^{\perp }$ under $B$, it also annihilates $E$, and hence all of $V$; nondegeneracy makes it zero. The restricted alternating bilinear form on $E^{\perp }$ is therefore again nondegenerate. Induction splits off two dimensions at each step and ends at the zero space. HenceThe proof works in characteristic two as well: alternation, rather than merely skew symmetry, is the needed hypothesis.