I know that Wronskian method is easy to apply in order to find if there is linear independence for a family of functions and also mathematical induction too is applicable as given in this post for a family of trigonometric functions of $\sin(kx)$ where $k \in \mathbb{N} - \{ 0 \}$ However, I could not figure out how to apply the induction for this problem to prove $S$ is linearly independent where $S = \{ f_{k}(x) = x^{k} : k \in \mathbb{N} \}$ for all $x$ except for $x = 0$ since indeterminate form $0^0$.
How can I prove $S$ is composed of linearly independent functions via mathematical induction?
Let's show by induction that for all $n \geq 0$, the family $\lbrace f_0, ..., f_n \rbrace$ is linearly independant.
This is true for $n=0$, since $f_0 \neq 0$. Let's suppose that it is true for an integer $n$. Then let's suppose that you have $a_0$, ..., $a_n$, $a_{n+1}$ real numbers such that $$a_0 f_0 + ... + a_n f_n + a_{n+1}f_{n+1} = 0 \quad \quad \quad (1)$$
Differentiating this equality (and because $f_0'=0$ and $f_k'=kf_{k-1}$) you get $$\sum_{k=0}^{n} (k+1)a_{k+1} f_{k}= 0$$
By the induction hypothesis, the family $\lbrace f_0, ..., f_n \rbrace$ is linearly independant, so you deduce that $(k+1)a_{k+1} = 0$ for all $k=0, ..., n$, i.e. that $a_{k+1} = 0$ for all $k=0, ..., n$. Injecting this information in the equation $(1)$, you deduce finally that $a_0= 0$. So all the $a_0, ..., a_{n+1}$ are zero, so the family $\lbrace f_0, ..., f_n, f_{n+1} \rbrace$ is linearly independant.
This completes the induction.