Here is a broader context for your question. In every finite Galois extension $L/K$ there is an $\alpha \in L$ such that the $K$-conjugates $\{\sigma(\alpha) : \sigma \in G\}$ form a $K$-basis of $L$, where $G$ is the Galois group. This is called a normal basis. As an example, in $\mathbf Q(i)/\mathbf Q$ the sets $\{1,i\}$ and $\{i,-i\}$ are not normal bases but $\{1+i,1-i\}$ is a normal basis. Here is the key result: when $H$ is a subgroup of $G$, its fixed field $L^H$ is $K(\alpha_H)$ where $\alpha_H = {\rm Tr}_{L/L^H}(\alpha) = \sum_{\tau \in H} \tau(\alpha)$. The proof of this uses linear independence of characters, as other answers have already done in the particular case you ask about. See Theorem 3.8 here.
You ask about $K = \mathbf Q$, $L = \mathbf Q(\zeta_p)$, $G = (\mathbf Z/p\mathbf Z)^\times$, $\alpha = \zeta_p$, and $H = \{y \bmod p : y^k \equiv 1 \bmod p\}$. That $\zeta_p$ and its $\mathbf Q$-conjugates are a normal basis of $\mathbf Q(\zeta_p)/\mathbf Q$ is Example 3.2 here. Watch out: when $n \geq 1$, $\zeta_n$ and its $\mathbf Q$-conjugates are not always a normal basis of $\mathbf Q(\zeta_n)/\mathbf Q$: they are if and only if $n$ is squarefree and that is shown in the top answer here.
