Denote by $x^s\equiv \prod_{k=1}^nx_k^{s_k}$ when $x\in\mathbb{R}^n, s\in\mathbb{N}_0^n$ and use the convention that $0^0 = 1$. Define a partial order for elements of $\mathbb{N}_0^n$ with $\alpha\leq \beta \Longleftrightarrow \exists 1\leq k\leq n:\alpha_k\leq \beta_k\land \forall l=1,\dots,k-1:\alpha_l = \beta_l$.
Denote by $|x|$ the Euclidean norm of an element of $\mathbb{R}^n$.
I am looking either for a hint or a full proof for the following: Let $N\in\mathbb{N}$. Then there exists $C_N > 0$ such that
$$C_N^{-1}(1+|x|)^N\leq\sum_{|s|\leq N}|x^s|\leq C_N(1+|x|)^N$$
A book I am reading states that this is an easy estimate, so this could result from a simple observation. Partly for this reason I have not made any progress with the proof since if I write everything in terms of $x$'s components, it is impossible for me to see the trick.
On the one hand, we have that $\prod_{i =1}^{k} |x_i|^{s_i} \le |x|^{|s|}$ where $|s| = \sum_{i=1}^n s_i$ so that $$\sum_{|s| \le N} |x^s| \lesssim \sum_{j \le N} |x|^j \le \sum_{j \le N} \frac{N!}{j! (N-j)!}|x|^j = (1+|x|)^N.$$
On the other hand, $|x| \lesssim \max_i |x_i|$ so that $|x|^k \lesssim |x^{\alpha(k,x)}|$ where $\alpha(k,x)$ is the multi-index that is $k$ in the $\arg \max_i |x_i|$ component and $0$ in other components. As a result, $$(1+|x|)^N = \sum_{j \le N}\frac{N!}{j! (N-j)!} |x|^j \lesssim \sum_{j \le N} |x|^j \lesssim \sum_{j \le N} |x^{\alpha(j,x)}| \le \sum_{|s| \le N} |x^s|.$$