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Theorem: ${f \in \Bbb{Z}[X]}$ is irreducible, monic and all roots of ${f}$ have absolute value at most 1. Then $f$ is a cyclotomic polynomial.

Here is a part of the proof I feel confused: It suffices to prove that every root of $f$ is a root of unity. If $f(X)=(X-\alpha_1)...(X-\alpha_r)$consider the family of polynomials $f_n(X) =(X-\alpha_1^n)...(X-\alpha_r^n)$.The coefficients of ${f_n}$ are algebraic integers, since they are calculated using multiplications and additions starting from ${\alpha_1,...,\alpha_r}$. On the other hand, the coefficients of ${f_n}$ are symmetric polynomials in ${(\alpha_i)}$, so they are rational, and therefore integers.

I don't understand why "the coefficients of ${f_n}$ are symmetric polynomials in ${(\alpha_i)}$, so they are rational, and therefore integers."? Why is it rational? And why therefore integer?

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As it is written right now, the claim is false. Indeed, $f=X$ is an irreducible monic polynomial with integers coefficients and whose roots lie in the unit disk. However, it is not a cyclotomic polynomial. The result will hold if you assume that $0$ is not a root of the given polynomial! Note that I am being picky, this is the only counterexample. In fact, one has:

Theorem. An irreducible monic polynomial with integers coefficients is either $X$ or a cyclotomic polynomial.

Regarding your question, notice that for $\ell\in\{0,\cdots,r\}$, the coefficient of $X^{r-\ell}$ in $f_n$ is: $$(-1)^\ell\sigma_{\ell}({\alpha_1}^n,\ldots,{\alpha_r}^n),$$ where $\sigma_{\ell}$ is the $\ell$-th elementary symmetric polynomial. Furtheremore, notice that $\sigma_{\ell}({X_1}^n,\ldots,{X_r}^n)$ is a symmetric polynomial and therefore there exists $S_{\ell}\in\mathbb{Z}[X_1,\cdots,X_n]$ such that: $$\sigma_{\ell}({X_1}^n,\ldots,{X_r}^n)=S_\ell(\sigma_1(X_1,\ldots,X_r),\ldots,\sigma_r(X_1,\ldots,X_r)).$$ In particular, one has: $$\sigma_{\ell}({\alpha_1}^n,\ldots,{\alpha_r}^n)=S_\ell(\sigma_1(\alpha_1,\ldots,\alpha_r),\ldots,\sigma_r(\alpha_1,\ldots,\alpha_r)).$$ However, for all $k\in\{0,\cdots,r\}$, $\sigma_k(\alpha_1,\cdots,\alpha_r)$ is itself an integer as it is the the coefficient of $X^{n-k}$ in $f\in\mathbb{Z}[X]$ multiply by $(-1)^r$. Whence the result, since $\mathbb{Z}$ is closed under addition and multiplication.

The result you are trying to prove is a consequence of:

Theorem. (Kronecker) Let $f$ be a monic polynomial with integers coefficients whose complex roots are nonzero and lie in the unit disk, then the roots of $f$ are roots of unity.

Remark. The key point is that if $A$ is a commutative ring with a units, then $A[X_1,\cdots,X_n]^{S_n}$ is a finitely generated $A$-module. Moreover, the elementary symmetric polynomials are generators: $$\sum_{\substack{I\subseteq\{0,\cdots,n\}\\|I|=k}}\prod_{i\in I}X_i$$

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    Do you agree that I have proved that $\sigma_{\ell}({\alpha_1}^n,\ldots,{\alpha_r}^n)$ is a $\mathbb{Z}$-linear combinations of $\sigma_k(\alpha_1,\ldots,\alpha_r)$? – 2017-01-30
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    I am going to expand this point in the main post, besides this is not exactly a $\mathbb{Z}$-linear combinations, powers are allowed. Thank you for your remark! – 2017-01-30
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    Ok, if $|\alpha_j| \le 1$ then the coefficients of $f_n$ are bounded, and since $f_n \in \mathbb{Z}[X]$ and have constant degree, it means $f_n = f_m$ for some $m \ne n$,. Thus, $\alpha_j = \alpha_i^{n-m}$, iterating more times we should get $\alpha_j = \alpha_j^{n-k}$ and $\alpha_j$ is a root of unity – 2017-01-30
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    @user1952009 Yes, that is exacly the point! The set of monic polynomials with integer coefficients and whose roots are nonzero and lie in the unit disk is finite! – 2017-01-30
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    For the generalization, it works only if $\{a \in A, |a| < c\}$ is finite for some norm $|.|$, the norm that we use for defining the unit disk containing the roots ? (with $A = \mathbb{Z}[\sqrt{5}]$ there is no such norm) – 2017-01-30