For $\mathbf{x} = (x_1, x_2, \cdots, x_\nu) \in \mathbb{R}^\nu$, where $\nu = \max \{ j(1), \cdots, j(n) \} $, define $f :\mathbb{R}^\nu \rightarrow \mathbb{R}^n$ as follows.
$$ f(\mathbf{x} ) = \left( \sum_{i=1}^{j(1)}x_i ,\sum_{i=1}^{j(2)} x_i^2, \sum_{i=1}^{j(3)} x_i^3, \cdots, \sum_{i=1}^{j(n)}x_i^n \right) $$
Say $\exists \mathbf{c} = (c_1, \cdots, c_n) \in \mathbb{R}^n$ such that $\mathbf{c}^Tf(\mathbf{x}) \equiv 0$. I'm trying to show $\mathbf{c} = \mathbf{0}$, uniquely.
What comes to my mind is uniqueness of power series, but I'm not sure it can be thought of as "uniqueness of polynomial expansion".
P.S. maybe the problem can be trimmed more tidily, but I have no idea for now.
If:
$$ f(\mathbf{x} ) = \left( \sum_{i=1}^{j(1)}x_i ,\sum_{i=1}^{j(2)} x_i^2, \sum_{i=1}^{j(3)} x_i^3, \cdots, \sum_{i=1}^{j(n)}x_i^n \right) $$
Then:
$$ c^T f(\mathbf{x} ) = c_1\sum_{i=1}^{j(1)}x_i + c_2\sum_{i=1}^{j(2)} x_i^2 + c_3\sum_{i=1}^{j(3)} x_i^3 + \cdots + c_n\sum_{i=1}^{j(n)}x_i^n $$
Then if:
$$ c^T f(\mathbf{x} ) \equiv 0 $$
For any $i$, $j$ we have: $$ \frac{\partial ^i [c^T f(\mathbf{x} )]}{\partial x_j^i} \equiv 0 $$
For $i=n$: $$ \frac{\partial ^n [c^T f(\mathbf{x} )]}{\partial x_j^n} = c_n\sum_{i=j}^{j} n! \equiv 0 \Rightarrow c_n=0 $$
Then we eliminate the last term in $c^T f(\mathbf{x} )$. By repeating until $n=1$ we conclude that $c$ is a null vector.