The Mahler measure of a integral polynomial is an algebraic integer

Notice first that the Mahler measure is multiplicative, hence it suffices to consider an irreducible polynomial. Namely, fixing $\alpha \in \overline{\Bbb Q}$, and letting $f \in \Bbb Z[X]$ be the minimal integer (in general non monic) polynomial of $\alpha$, it suffices to consider $M(f)$ in this context. Then, according to Proposition 1.6.6 of Heights in Diophantine Geometry (Bombieri, Gubler): $$M(f) = H(\alpha)^{\deg \alpha}$$ where $H(\alpha)$ is the multiplicative height of $\alpha$. Let me recall what this is.

Let $K = \Bbb Q (\alpha)$, and set $P_K$ the set of finite places of $K$. Then, for $v \in P_K$ above a prime $p_v$, and $x$ in the completion $K_v$, set $$|x|_v = |N_{K_v/\Bbb Q_{p_v}}(x)|_{p_v}^{\frac{1}{[K:\Bbb Q]}} = |N_{K_v/\Bbb Q_{p_v}}(x)|_{p_v}^{\frac{1}{\deg \alpha}}.$$ Then, $H(\alpha)$ is defined as $$H(\alpha) = \prod_{v \in P_K} \max(1, |\alpha|_v)$$ and so $$H(\alpha)^{\deg \alpha} = \prod_{v \in P_K} \max(1, |N_{K_v/\Bbb Q_{p_v}}(x)|_{p_v}).$$ I claim that this product is an algebraic integer. Indeed, every term on the right hand side is of the form $p_v^{a/b}$ with $a \ge 0$ (because of the maximum) and $b \ge 1$, which is a root of the monic, integer coefficient polynomial $X^b-p^a$. This concludes the argument.

There is a much easier proof to show that $|a_n|^{n-1}M(f)$ is an algebraic integer. Write: $$f(X)=\sum_{i=0}^n a_i X^i=a_n\prod_{j=1}^n(X-\alpha_j) \in \Bbb Z[X].$$ Multiply by $a_n^{n-1}$: $$f(X)=\sum_{i=0}^n a_ia_n^{n-1-i}(a_nX)^i=\prod_{j=1}^n((a_nX)-a_n\alpha_j).$$ Set: $$g(Y)=Y^n+\sum_{i=0}^{n-1} a_ia_n^{n-1-i}Y^i=\prod_{j=1}^n(Y-a_n\alpha_j)$$ which is a monic polynomial with integer coefficients, and roots $a_n\alpha_j$. This shows that the $a_n\alpha_j$ are algebraic integers. Moreover, the complex norm of an algebraic integer is again an algebraic integer. Hence $$|a_n|^{n-1}M(f)=\prod_{j=1}^{n} \max(|a_n|,|a_n \alpha_j|)$$ is a product of algebraic integers, so is an algebraic integer.