= 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 $V\ne0$, choose $v\ne0$. Nondegeneracy supplies $w$ with $B(v,w)\ne0$; rescale $w$ so that $B(v,w)=1$. The vectors are independent, and the restriction to $E=\operatorname{span}\{v,w\}$ has matrix $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$, 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. Hence
$$
\boxed{\dim V\text{ is even}.}
$$
The proof works in characteristic two as well: alternation, rather than merely skew symmetry, is the needed hypothesis.
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