Let $p$ be a prime, and consider the finite field $\mathbb{F}_p$. Fix any $n\ge1$, and consider the field extension $\mathbb{F}_{p^n}/\mathbb{F}_p$. If $\alpha\in\mathbb{F}_{p^n}$ is a multiplicative generator, then $\alpha$ is also a primitive element, that is, $\mathbb{F}_{p^n}=\mathbb{F}_p(\alpha)$, and further, the set $\{1,\alpha,\ldots,\alpha^{n-1}\}$ is an $\mathbb{F}_p$-linear basis of $\mathbb{F}_{p^n}$. Therefore, we can conclude that the elements $\alpha,\ldots,\alpha^{n-1}\not\in\mathbb{F}_p$.
In the above statement, I could conclude 'non-membership' in $\mathbb{F}_p$ by using linear independence over $\mathbb{F}_p$. Is a converse of this true, that is, must 'non-membership' imply linear independence over $\mathbb{F}_p$? Specifically, for $n,m\ge1$, if $\alpha\in\mathbb{F}_{p^n}$ is a multiplicative generator such that the elements $\alpha,\ldots,\alpha^{m-1}\not\in\mathbb{F}_p$, then must $m\le n$?
Let $p=2$ and $n=3$, and let $\alpha$ be a generator of the multiplicative group, so that $\alpha$ has order $7$.
Then none of $\alpha,\alpha^2,\alpha^3,\alpha^4,\alpha^5,\alpha^6$ lie in $\mathbb{F}_2$. Yet $7>3$.