[guided]The final step is pure bookkeeping. Expand the factorisation $(\star)$:
\begin{align*}
\chi_\beta(x) = \prod_{i=1}^n (x - \sigma_i(\beta)).
\end{align*}
Using Vieta's formulas (or, equivalently, direct expansion and matching coefficients):
\begin{align*}
\chi_\beta(x) &= x^n - \left( \sum_{i=1}^n \sigma_i(\beta) \right) x^{n-1} + \left(\sum_{i < j} \sigma_i(\beta)\sigma_j(\beta)\right) x^{n-2} - \cdots + (-1)^n \prod_{i=1}^n \sigma_i(\beta).
\end{align*}
Writing $\chi_\beta(x) = x^n + c_{n-1} x^{n-1} + \cdots + c_0$, we read off:
\begin{align*}
c_{n-1} &= -\sum_{i=1}^n \sigma_i(\beta), \\
c_0 &= (-1)^n \prod_{i=1}^n \sigma_i(\beta).
\end{align*}
Substituting into $(\ast)$ from the first step:
\begin{align*}
\operatorname{tr}_{L/\mathbb{Q}}(\beta) = -c_{n-1} = -\left(-\sum_{i=1}^n \sigma_i(\beta)\right) = \sum_{i=1}^n \sigma_i(\beta).
\end{align*}
And:
\begin{align*}
N_{L/\mathbb{Q}}(\beta) = (-1)^n c_0 = (-1)^n \cdot (-1)^n \prod_{i=1}^n \sigma_i(\beta) = \prod_{i=1}^n \sigma_i(\beta),
\end{align*}
where we used $(-1)^n \cdot (-1)^n = (-1)^{2n} = 1$.
This gives the desired formulas:
\begin{align*}
\operatorname{tr}_{L/\mathbb{Q}}(\beta) = \sum_{i=1}^n \sigma_i(\beta), \qquad N_{L/\mathbb{Q}}(\beta) = \prod_{i=1}^n \sigma_i(\beta),
\end{align*}
and the values $\sigma_1(\beta), \ldots, \sigma_n(\beta)$ are the **conjugates** of $\beta$ in $\mathbb{C}$, partitioned as $\{\beta_1, \ldots, \beta_1, \beta_2, \ldots, \beta_2, \ldots, \beta_m, \ldots, \beta_m\}$ with each $\beta_j$ repeated $r$ times.
**A structural remark**: when $\beta$ is a primitive element, i.e. $L = \mathbb{Q}(\beta)$, then $r = 1$ and $m = n$, and the conjugates are simply the $n$ distinct roots of $p_\beta$ in $\mathbb{C}$. When $\beta$ is a "sub-primitive" element generating a proper subfield $\mathbb{Q}(\beta) \subsetneq L$, then $r > 1$ and the conjugates of $\beta$ come with repetitions: each distinct value $\beta_j$ appears exactly $r$ times in the list $(\sigma_1(\beta), \ldots, \sigma_n(\beta))$. The trace and norm formulas automatically take this multiplicity into account because they sum/multiply over all $n$ embeddings, not over the distinct roots of $p_\beta$.[/guided]