Prove the linear independence of $\{e^{ix} x^j\}_{i,j}$ I apologize in advance if this question has been asked before, but I am unable to find it. How would one prove $\{e^{ix} x^j\}_{i,j\in \mathbb Z},\,x\in \mathbf R$ is linearly independent?
 A: We use induction in the number of distinct terms in the liner combination
$$c_1e^{n_1t}t^{m_1}+\ldots + c_ke^{n_kt}t^{m_k}=0$$
For $k=1$ the statement holds trivially.
Suppose the any  $1\leq j<k$ distinct terms of the form $t^me^{nt}$ are linearly independent.  Consider the sum
$$c_1t^{m_1}e^{n_1t}+c_2e^{n_2t}t^{m_2}+\ldots+c_ke^{n_kt}t^{m_k}=0$$
Suppose the terms in the sum above have been ordered in such a way that $m_1\leq m_2\leq \ldots \leq m_k$.
Dividing by $t^{m_1}$ we get that
$$\begin{align}
c_1e^{n_1t}+c_2e^{n_2t}t^{m_1-m_2}+\ldots+c_ke^{n_kt}t^{m_k-m_1}=0\tag{1}\label{one}
\end{align}$$

*

*If all $n_j$, $1\leq j\leq k$, are the same, then the $m_j$ are all distinct and
$$c_1+c_2t^{m_2-m_2}+\ldots + t^{m_k-m_1}=0$$
for all $t$, which means that $c_1=\ldots=c_k=0$ by the fundamental theorem of algebra.


*If not all the $n_j$'s are the same, then we can rearrange the terms in \eqref{one} as
$$p_1(t)e^{n'_1t}+\ldots+ p_\ell(t)e^{n'_\ell t}=0$$
so that $p_1,\ldots, p_\ell$  are polynomials and $n'_1<\ldots< n'_\ell$. Hence
$$p_1(t)e^{(n'_1-n'_\ell)t}+\ldots + p_{\ell-1}(t)e^{(n'_{\ell-1}-n_\ell)t}+  p_\ell(t)=0$$
for all $t$. Letting $t\rightarrow\infty$ yields
$$0=\lim_{t\rightarrow\infty}|p_1(t)e^{(n'_1-n'_\ell)t}+\ldots + p_{\ell-1}(t)e^{(n'_{\ell-1}-n_\ell)t}+  p_\ell(t)|=\lim_{t\rightarrow\infty}|p_\ell(t)|$$
Hence $p_\ell(t)=0$ (otherwise $0=\lim_{t\rightarrow\infty}|p_\ell(t)|>0$ which is absurd.
This reduced the terms number of terms in \eqref{one} and we can then apply the induction hypothesis.
