Sum and product of all linear combinations

Let $$K$$ be a compact non-empty subset of $$\mathbb{R}^N$$. If $$x \in \mathbb{R}^N$$ and if $$(k_1, k_2, \dots, k_N), k_i \in \mathbb{Z}_+$$, consider the function $$x \mapsto x_1^{k_1}x_2^{k_2}...x_N^{k_N}.$$

I want to show that the set $$\mathcal{A}$$ of all linear combinations of such functions is a $$\mathbb{R}$$-subalgebra of $$C(K)$$, but I'm struggling specifically with proving that the sum and the product of elements of $$\mathcal{A}$$ belongs to $$\mathcal{A}$$ cause I've very much confused with manipulating the indexes.

$$\textbf{My attempt:}$$ Consider the set $$\mathcal{A} = \Big\{ \sum_{i=1}^{N}\alpha_i\prod_{i=1}^{N} x_i^{k_i}: x_i, \alpha_i \in \mathbb{R}, k_i \in \mathbb{Z}_+ \Big\}$$ 1. $$p, q \in \mathcal{A} \Rightarrow pq \in \mathcal{A}$$

Take $$p, q \in \mathcal{A}$$ such that $$p = \sum_{i=1}^{N}\alpha_i\prod_{i=1}^{N} x_i^{k_i}$$ and $$q = \sum_{j=1}^{N}\tilde{\alpha}_j\prod_{j=1}^{N} x_j^{\tilde{k}_j}$$, then it follows:

\begin{align*} p(x)q(x) &= \Big(\sum_{i=1}^{N}\alpha_i\prod_{i=1}^{N} x_i^{k_i} \Big) \Big(\sum_{j=1}^{N}\tilde{\alpha_j}\prod_{j=1}^{N} x_j^{\tilde{k}_j} \Big) \\ &= \sum_{l=1}^{2N}c_l \prod_{i=1}^{l}x_i^{k_i}, \\ &= (pq)(x) \end{align*} with $$c_l = \sum_{m=1}^{l}\alpha_m\tilde{\alpha}_{l-m}$$.

1. $$p, q \in \mathcal{A} \Rightarrow p+q \in \mathcal{A}$$

Take the same $$p, q \in \mathcal{A}$$ written above, then \begin{align*} p(x) + q(x) &= \Big(\sum_{i=1}^{N}\alpha_i\prod_{i=1}^{N} x_i^{k_i} \Big) + \Big(\sum_{j=1}^{N}\tilde{\alpha_j}\prod_{j=1}^{N} x_j^{\tilde{k}_j} \Big) \\ &= \sum_{l=1}^{N} (\alpha_l + \tilde{\alpha}_l)\prod_{l=1}^{N}x_l^{k_l} \\ &= (p + q)(x) \end{align*}