Show that $n$ is accumulation points of this complex sequence 
Let $z_1,...,z_n \in \mathbb C$ such that  $\displaystyle  \forall i  |z_{i}|= 1$. Denote $\displaystyle u_{k}= \sum_{i=1}^{n} z_{i}^{k}$.
Show that $n$ is accumulation points$\color{red}{^{[1]}}$ of $(u_{k})$.

My attempt :
Let $z_j=e^{i \theta_j}$,
then,
$|u_k-n| \leq \sum_{j=1}^n |1-e^{ik \theta_j}| = \sum_{j=1}^n |\int_0^{k \theta_j - 2\pi \lfloor \frac{k \theta_j}{2\pi} \rfloor} e^{it}dt|\leq 2\pi \sum_{j=1}^n k \alpha_j - \lfloor k \alpha_j \rfloor$
Afer this I don't how can I continue,
Thank you in advance,

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*$\color{red}{[1]}$ $\forall \varepsilon >0,\forall N\in\mathbb{N},  (\exists n\ge N, \;|u_n-y|<\varepsilon)$
 A: This approach depends on Kronecker's theorem (see Trajectories on the $k$-dimensional torus for a reference to a proof). A more straightforward answer escapes me at present.
Let $\mathbb{T}^n = \mathbb{R}^n / \mathbb{Z}^n$. For $x\in \mathbb{R}^n]$, let $[x] = \{x\} + \mathbb{Z}^n$, that is, the equivalence classes under the equivalence relation $x \sim y$ iff $x-y \in \mathbb{Z}^n$.
Note that the estimate in the question reduces to showing that the sequence $k \mapsto [\sum_{j=1}^n k \alpha_j ]$ has an accumulation point at $[0] \in \mathbb{T} $.
First suppose that the $\alpha_j$ are linearly independent over $\mathbb{Q}$, and let
$\alpha = (\alpha_1,...,\alpha_n)^T$. (Note that if the $\alpha_j$ are linearly independent over $\mathbb{Q}$, then so are $q \alpha_j$ for any non-zero $q \in \mathbb{q}$.)
Then the above result shows that the trajectory $k \mapsto [k \alpha]$ is dense in $\mathbb{T}^n $. In particular, we can find a sequence $k_i$ such that
$[k_i \alpha] \to [0]$. Then the estimate in the question shows that $u_{k_i} \to n$.
Now we deal with the general case.
By reordering if necessary, suppose $\alpha_1,..., \alpha_l $ are a basis for
$\operatorname{sp}_\mathbb{Q} \{ \alpha_1,...,\alpha_n \}$ (that is, the span over 
$\mathbb{Q}$), with $l<n$. Then we can write, for $j >l$,
$\alpha_i = \gamma_j^T \bar{\alpha}$, where $\gamma_j \in \mathbb{Q}^{n-l}$, and
$\bar{\alpha} = ( \alpha_1,..., \alpha_l )^T$.
By choosing $p \in \mathbb{Z}$ appropriately, we can write $\gamma_j = {1 \over p} \lambda_j$, where $\lambda_j \in \mathbb{Z}^{n-l}$.
As above,  we can find a sequence $k_i$ such that $[k_i\bar{\alpha}] \to [0] \in \mathbb{T}^l$. In particular, this means there exists $n_i \in \mathbb{Z}^l$ such that
$k_i \bar{\alpha}+n_i \to 0$ (in $\mathbb{R}^l$). Then with $p \in \mathbb{Z}$, we see that $p(k_i \bar{\alpha}+n_i) \to 0$ and so$[pk_i\bar{\alpha}] \to [0]$ as well. Now note that $\gamma_i^T(p(k_i \bar{\alpha}+n_i)) = \gamma_i^T(pk_i \bar{\alpha})+\lambda_i^T(n_i) = \lambda_i^T (k_i \bar{\alpha}+n_i)$, from which we get $[\gamma_i^T(pk_i \bar{\alpha})] \to [0] \in \mathbb{T}$. Consequently, we see that $[p k_i \alpha] \to [0] \in \mathbb{T}^n$. As above, this shows that $u_{(p k_i)} \to n$.
A: I hope the following proof is correct (@copper.hat proof is faultless, but I just share a straightforward answer)


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*Let $v_k=(z_1^k,z_2^k,\cdots,z_n^k)\in \mathbb{C^n}$.


The sequence $(v_k)$ is bounded, then by Weierstrass theorem we can extract $v_{\phi(k)}$ wich converge to $(l_1,l_2,\cdots, l_n)$ (non-zero by hypothesis). 
Even if we reextracting. Assume that $\psi(k)=\phi(k+1)-\phi(k)$ is strictly increasing . 
As $\frac{z_i^{\phi(k+1)}}{z_i^{\phi(k)}}\rightarrow 1$ for all $i\in \{1,\cdots,n\}$.
Thus, $z_i^{\psi(k)}\rightarrow 1$.
Therefore $u_{\psi(k)}\rightarrow n$ $\square$
