[proofplan] We compare $P+P$ with the image of a larger integer box. Adding two elements of $P$ adds their coordinates, so the $i$th coordinate ranges from $0$ to $2L_i-2$. This gives an upper bound for $|P+P|$ by the size of that larger box, while properness identifies $|P|$ with the size of the original box. [/proofplan]

[step:Encode sums of two elements by a larger coordinate box] Define the original coordinate box \begin{align*} B := \{0,1,\dots,L_1-1\} \times \cdots \times \{0,1,\dots,L_d-1\}. \end{align*} Define the doubled coordinate box \begin{align*} C := \{0,1,\dots,2L_1-2\} \times \cdots \times \{0,1,\dots,2L_d-2\}. \end{align*} Define the map \begin{align*} \psi: C &\to G \\ (m_1,\dots,m_d) &\mapsto 2x_0 + \sum_{i=1}^d m_i x_i, \end{align*} where $2x_0$ denotes $x_0+x_0$. We claim that $P+P \subseteq \psi(C)$. Let $y \in P+P$. Then there exist $p,q \in P$ such that $y=p+q$. By the definition of $P$, there exist $(n_1,\dots,n_d),(k_1,\dots,k_d) \in B$ such that \begin{align*} p &= x_0 + \sum_{i=1}^d n_i x_i, & q &= x_0 + \sum_{i=1}^d k_i x_i. \end{align*} For each $i \in \{1,\dots,d\}$, define $m_i := n_i+k_i$. Since $0 \leq n_i,k_i \leq L_i-1$, we have \begin{align*} 0 \leq m_i \leq 2L_i-2. \end{align*} Hence $(m_1,\dots,m_d) \in C$. Using commutativity in $G$, \begin{align*} y &= p+q \\ &= \left(x_0 + \sum_{i=1}^d n_i x_i\right) + \left(x_0 + \sum_{i=1}^d k_i x_i\right) \\ &= 2x_0 + \sum_{i=1}^d (n_i+k_i)x_i \\ &= 2x_0 + \sum_{i=1}^d m_i x_i \\ &= \psi(m_1,\dots,m_d). \end{align*} Therefore $P+P \subseteq \psi(C)$. [/step]

[step:Bound the size of $P+P$ by counting the larger box] Since $P+P \subseteq \psi(C)$, monotonicity of cardinality for finite sets gives \begin{align*} |P+P| \leq |\psi(C)|. \end{align*} The image of a finite set has cardinality at most the cardinality of its domain, so \begin{align*} |\psi(C)| \leq |C|. \end{align*} By the product rule for finite Cartesian products, \begin{align*} |C| = \prod_{i=1}^d |\{0,1,\dots,2L_i-2\}| = \prod_{i=1}^d (2L_i-1). \end{align*} Combining these inequalities gives \begin{align*} |P+P| \leq \prod_{i=1}^d (2L_i-1). \end{align*} [/step]

[step:Use properness to convert the box count into $|P|$] Because $P$ is proper, the parametrization \begin{align*} \phi: B &\to P \\ (n_1,\dots,n_d) &\mapsto x_0 + \sum_{i=1}^d n_i x_i \end{align*} is injective. It is also surjective by the definition of $P$. Hence $\phi$ is a bijection from $B$ onto $P$, and therefore \begin{align*} |P| = |B| = \prod_{i=1}^d L_i. \end{align*} For each $i \in \{1,\dots,d\}$, the inequality $2L_i-1 \leq 2L_i$ holds because $L_i \geq 1$. Therefore \begin{align*} \prod_{i=1}^d (2L_i-1) \leq \prod_{i=1}^d 2L_i = 2^d \prod_{i=1}^d L_i = 2^d |P|. \end{align*} Together with the previous step, this proves \begin{align*} |P+P| \leq 2^d |P|. \end{align*} [/step]