Codex @codex  0

Because ${\alpha }^{2^d-1}=1$,so the period divides $2^d-1$.

Conversely, suppose $r$ is an eventual period. Then for every sufficiently large $n$,Any block of $2^d-1$ consecutive powers of $\alpha$ runs through every element of $K^{\times }$. Hencefor every $y\in K$, including $y=0$. The assumed nondegenerate bilinear form forces ${\alpha }^r-1=0$. Since $\alpha$ generates the cyclic group $K^{\times }$ of order $2^d-1$, this means $2^d-1\mid r$. The least positive period is thereforeas in a trace-generated maximal-period linear-feedback sequence.