Linear dependence of $\left\{x^{n}\,\colon\, n\in\mathbb{N}\right\}$ Consider the set $S=\left\{x^{n}\,\colon\, n\in\mathbb{N}\right\}$. (Note that $x\in\mathbb{R}$) Is this set linearly dependent? 
Well thinking about it we want to find some non-trivial values $\lambda_{n}$ such that $$\lambda_{1}x+\lambda_{2}x^{2} + \ldots + \lambda_{n}x^{n} + \ldots = 0.$$ In other words we want to find if $$\sum_{k=1}^{\infty}\lambda_{n}x^{n} = 0$$ for any $\lambda_{n}\not=0$. However i'm not really sure how to proceed.
 A: An approach that doesn't involve differentiation: it suffices to show that
$$
f(x) = \lambda_{1}x+\lambda_{2}x^{2} + \cdots + \lambda_{n}x^{n} = 0 \implies \lambda_i = 0
$$
Note that if any of the $\lambda_i$ is non-zero, $\lim_{x \to \infty} f(x)$ is either $\infty$ or $-\infty$, depending on the sign of the coefficient of the non-zero term of greatest degree.  If $|\lim_{x \to \infty} f(x)| = \infty$, then $f(x) \neq 0$.
Thus, if any of the $\lambda_i$ are non-zero, then $f(x) \neq 0$. By contrapositive, the conclusion follows.
A: You are probably considering the set as a subset of vector space of real functions on $\mathbb{R}$ (or continuous functions, perhaps, but it's the same).
A set is linearly independent if and only if any finite subset is linearly independent, and so we can simply show that
$$
\{1,x,x^2,\dots,x^n\}
$$
is linearly independent. Suppose $\lambda_0+\lambda_1x+\dots+\lambda_nx^n=0$ (that is, the constant zero function); this is a polynomial that has infinitely many roots, so it's the zero polynomial, which means $\lambda_0=\lambda_1=\dots=\lambda_n=0$.
A: I presume you are talking about the functions $b_n(x) = x^n$.
Suppose you have $\alpha_k$ ($k=1,..,N$) such that
$f(x)=\sum_k \alpha_k b_k(x) = 0$ for all $x$. Note that
$f^{(k)}(0) = 0$ for all $k \in \mathbb{N}$.
It is not hard to show that $f^{(k)}(0) = k! \alpha_k $, from which it follows that
$\alpha_k = 0$. Hence the $b_n$ are linearly independent.
A: Let $\alpha_1<\ldots<\alpha_k\in\Bbb N$ and let $\lambda_1,\ldots,\lambda_k\in \Bbb R$ such that
$$P(x)=\lambda_1 x^{\alpha_1}+\cdots+\lambda_k x^{\alpha_k}=0$$
Now we have successively  $P^{(\alpha_i)}(x)=(\alpha_i)!\lambda_i=0$ for $i=\alpha_k,\ldots,\alpha_1$. Conclude.
