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Post Made Community Wiki by Fred Goodman
1

Kevin Buzzard gave the solution. Here it is with a little more detail:

Our assumptions include $\omega_0 = \omega_0^2$. Thus $\omega_0 \in \{0, 1\}$.

The linear homogeneous recursion only kicks in eventually; say the $\omega_a$ for $a \ge N$ satisfy such a recursion.

Let $v_1, \dots, v_m$ be the distinct roots of the characteristic polynomial of the linear recursion. Then there exist polynomials $h_1, \dots, h_m$ such that $\omega_a = \sum_{i = 1} ^m h_i(a) v_i^a$ for $a \ge N$. Let $\alpha_i$ be the constant term of $h_i$ for each $i$. Since the characteristic is $2$, we have $h_i(2a) = \alpha_i$ for all $a$.
For $a \ge N$,
$$\sum_i \alpha_i v_i^{4a} = \omega_{4 a} = \omega_{2a}^2 = \sum_i \alpha_i^2 v_i^{4a}.$$ Because the characteristic of $F$ is $2$, each element has a unique $2^k$--th root for all $k \ge 1$; in particular all the $v_i^4$ are distinct, so the displayed equation implies that $\alpha_i^2 = \alpha_i$ for all $i$, i.e. $\alpha_i \in \{0, 1\}$. Let $u_1, \dots, u_d$ be the list of those $v_j$ such that $\alpha_j = 1$. Then we have $\omega_{2a} = \sum_i u_i^{2a}$ for $a \ge N$. For an arbitrary $a \ge 1$, chose $k$ such that $2^{k-1} a \ge N$. Then $\omega_a$ is the unique $2^k$--th root of $\omega_{2^k a} = \sum_i u_i^{2^k a}$, namely $\omega_a = \sum_i u_i^a$.

Thus we have $\omega_0 \in \{0, 1\}$ and $\omega_a = p_a(u_1, \dots, u_d)$ for $a \ge 1$.

THANKS, KEVIN !