While proving that the character values in $\mathbb{C}$ of a finite group are algebraic integers, one usually proceeds as follows:
(1) Prove a very general theorem in algebraic number theory - algebraic integers are closed under addition and multiplication.
(2) For any character $\chi$ of a finite group $G$, $\chi(g)$ is sum of certain roots of unity in $\mathbb{C}$ (depending on $o(g)$).
(3) Since roots of unity are algebraic integers, so is their sum.
Question: Since the character values are sums of very specific algebraic integers: those $\alpha \in\mathbb{C}$ satisfying $\alpha^n=1$ for some $n$. Is there any direct and simple argument to prove that if $\alpha_1^{m_1}=\alpha_2^{m_2}=\cdots=\alpha_k^{m_k}=1$ then $\alpha_1+\cdots + \alpha_k$ is an algebraic integer? (We can assume that $m_1=m_2=\cdots=m_k$). In other words, is there way to avoid the very general statement in (1)?
Let us assume that all the $\alpha_j$ are $m$-th roots of unity. So if we fix some primitive $m$-th root of unity $\zeta$ then $\beta=\alpha_1+\cdots+\alpha_k=\zeta^{a_1}+\cdots+\zeta^{a_n}$ for some integers $a_j\ge0$. So we might as well take $\beta$ to be $f(\zeta)$ where $f$ is a polynomial with integer coefficients.
We have a matrix equation $Mv=\zeta v$ where $$M=\left(\begin{array}{ccccc} 0&1&0&\cdots&0\\ 0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 1&0&0&\cdots&0 \end{array}\right)$$ is an $n$-by-$n$ "circulant" matrix, and $v=(1\ \zeta\ \zeta^2\cdots\zeta^{n-1})^T$. It follows that $M^kv=\zeta^k v$ and so $f(M)v= f(\zeta)v =\beta v$. Thus $\beta$ is an eigenvalue of a matrix $N=f(M)$ which has integer coefficients. The characteristic equation of $N$ has integer coefficients, so $\beta$ is an algebraic integer.