To show that if all roots (complex) of an integer polynomial have norm 1, then they are roots of 1 We have, $f(x)=x^n + a_{n-1}x^{n-1} + \cdots + a_0 \in \mathbb{Z}[x]$, with $\alpha_i\in\mathbb{C},\ 1\leq i\leq n$ being all the roots of $f(x)$. If we have $|\alpha_i|=1$, for every $i$, then  $\alpha_i$ is a root of 1. 
Edit: As discussed in comments, I seem to have interpreted the problem incorrectly the first time around. 
Essentially, we have to show that ${\alpha_i}^k =1$, for some $k>n$.
Note: This question was posed after a class on Units, including Dirichlet's Unit Theorem. 
So I need some hint to help me get to the answer
 A: Let us collect the comments into an answer. By Newton's identities the power sums
$$ p(k) = \alpha_1^k+\ldots+\alpha_n^k $$
are integer numbers, and they belong to the $[-n,n]$ interval by the triangle inequality.
If $\alpha$ is a root of $f(x)$ then $\overline{\alpha}=\alpha^{-1}$ is also a root of $f(x)$ and by letting $\theta_i=\text{Arg}(\alpha_i)$ we have
$$ \cos(k\theta_1)+\ldots+\cos(k\theta_n)\in [-n,n]\cap\mathbb{Z} $$
for any $k\in\mathbb{N}$. If all the angles $\theta_i$ are rational multiples of $\pi$ there is nothing to prove.
Let us assume that $\theta_1\not\in\pi\mathbb{Q}$. In such a case, by the Lindemann-Weierstrass theorem we have that $\cos(k\theta_1)$ is a trascendental number over $\mathbb{Q}$ for any $k\in\mathbb{N}^+$. On the other hand $p(k)$ may only take a finite number of values, so for some $j\in[-n,n]$ there are infinite $k\in\mathbb{N}$ such that $p(k)=j$. Additionally $\cos(k\theta)=T_k(\cos\theta)$, hence by elimination of variables we get that some algebraic combination (with coefficients in $\mathbb{Q}$) of $\cos(k_1\theta_1),\cos(k_2\theta_1),\ldots,\cos(k_M \theta_1)$ equals zero. This contradicts the fact that $\cos(\theta_1)$ is trascendental over $\mathbb{Q}$, proving the claim.
A: Let me reformulate your question for security : If $\alpha$ is an algebraic integer all of whose conjugates have absolute value $1$, then $\alpha$ is a root of unity. Right ? Let us consider all the powers $\alpha^k$ for $k\in \mathbf N$. Such a power $\alpha^k$ is an algebraic integer, whose irreducible polynomial $f_k \in \mathbf Z[X]$ has degree less than the degree $n$ of $\alpha$ over $\mathbf Q$. But these coefficients are symmetric functions of the conjugates of $\alpha^k$, hence by the hypothesis and the triangle inequality, they are bounded by bounds depending only on $n$. It follows that there are only a finite number of possible $f_k$'s when $k$ varies. Therefore there are only finitely many distinct powers of $\alpha$, hence, after simplification, one of these powers must be $1$ .
A: We may suppose that all $\alpha_j$ are distincts. Put $b_k=\alpha_1^k+...+\alpha_n^k$. Then the $b_k$ are in $\mathbb{Z}$, as noticed by @Jack d'Aurizio, and clearly $|b_k|\leq n$ for all $k$. Now writing that the $\alpha_j$ are roots of $f$, multiplying by $\alpha_j^k$ and summing, we get that $$b_{k+n}+a_{n-1}b_{k+n-1}+...+a_0b_k=0$$
 for all $k$. Put $w_k=(b_k,b_{k+1},..,b_{k+n-1})$. By the above, the $w_k$ take only a finite number of values. Hence there exists $m$ and $h\geq 1$ such that $w_m=w_{m+h}$. This imply by the recurrence relation that we have $b_{k}=b_{k+h}$ for all $k\geq m$. Hence for all $k$ we have 
 $$(\alpha_1^h-1)\alpha_1^m\alpha_1^k+...+(\alpha_n^h-1)\alpha_n^m\alpha_n^k=0$$
Writing this for $k=0,...,n-1$ gives a linear system in the unknowns $x_j=(\alpha_j^h-1)\alpha_1^m$, with for determinant a Van der Monde determinant, hence non zero as the $\alpha_j$ are distincts. This imply that the $x_j$ are all $0$, and it is easy to finish. 
A: For $k$ natural number  consider the monic polynomial $f_k$ with the roots $\alpha_i^k$, $i=1, \ldots, n$. The coefficients of $f_k$ are integers (as integral polynomials in the coefficients of $f$) and bounded in absolute value ( $\le \binom{n}{ [\frac{n}{2}]}$ ), since all the $\alpha_i$ have absolute value $1$. Therefore, there exist $0\le k < l $ so that $f_k=f_l$. This means that the roots of these polynomials are equal, that, is, there exists a permutation $\sigma$ of $\{1, \ldots, n\}$ so that
$$\alpha_i^k= \alpha_{\sigma(i)}^l$$ for $i=1, \ldots, n$. From here we get $\alpha_i^{km}= \alpha_{\sigma(i)}^{lm}$, for all $m$, and hence, by induction on $t$
$$\alpha_i^{k^tm}= \alpha_{\sigma^t(i)}^{l^t m}$$ for all possible $i$, $m$, $t$.  Take now $t>1$ so that $\sigma^t=1$. We get
$$\alpha_i^{k^t}= \alpha_{i}^{l^t }$$ and so $\alpha_i^{l^t-k^t}=1$ for all $i$.
