= 1\}$$ where $$\ell(x_1,\dots,x_n) = \sum_{i = 1}^{(n-1)/2} x_i - \sum_{i = (n+1)/2}^n x_i,$$ and all other vertices $\mathbf{v}$ of $P_N$ satisfy $\ell(\mathbf{v}) = -1$. Moreover, the first $n$ vertices defining $F_0$ are clearly affinely independent, so $\dim F_0 = n-1$. Therefore, $F_0$ is indeed a facet of $P_N$.

Any other choice of $(\lambda_1,\dots,\lambda_n) \in \Lambda_n$ for the elements in $\mathcal{F}(P_N)$ will result in a facet as well, since the resulting convex hull is unimodularly equivalent to $F_0$. Hence, the same arguments can be applied to these sets. It remains to show that no other set of vertices will form a facet.

Take any element of  such that there are $k \geq 2$ more negative coefficients on the summands $\mathbf{e}_1,\dots,\mathbf{e}_n$ than positive coefficients, and set $\lambda_{n+1} = -1$. Without loss of generality, we can assume $\lambda_1 = \dots = \lambda_{(n-2k+1)/2} = 1$ and the remaining $\lambda_i = -1$. Call their convex hull $F'$. Form $\ell(x)$ as before and note that the first $n$ vertices of $F'$ satisfy $\ell(x_1,\dots,x_n) = 1$. Additionally, the vertices $\mathbf{v}$ of $P_N$ not in $F'$ satisfy $\ell(x_1,\dots,x_n) < 1$. However, $\ell(-\mathbf{e}_1 - \cdots - \mathbf{e}_n) = k$, so $\operatorname{aff}(F')$ actually passes through the interior of $P_N$ 