Is {$f_0, f_1,\dots,f_n$} a linearly independent set? Here is the question:  
Let $F$ be an infinite field and let c$_0$, c$_1$, ..., c$_n$ $\in$ $F$ be distinct. Then for $i$ = 0 ,1, ..., n, define
$f_i(x) =  \prod^n_{k=0,k\neq i}\frac{x-c_k}{c_i-c_k}$
Is {$f_0, f_1,...,f_n$} then a linearly independent set? Why?
So I tried to find a linear combination of {$f_0, f_1,...,f_n$}, namely $\exists$ $a_0, a_1, ..., a_n \in F$ and $a_0f_0+a_1f_1+...+a_nf_n=0$
I got the part that $\forall$ $i$, $f_i(c_i) = 1$ and $\forall$ $j\neq i, f_i(c_j)=0$
But then I figured that for $f_i(x)$, $x$ is arbitrary, then how is that if $x\neq c_i$?
Then take the linear combination, suppose $g=\sum^n_{i=0}a_if_i=0$ for some scalars $a_0, a_1,..., a_n$
Then, $g=\sum^n_{i=0}a_if_i(c_j)=0$ for $j=0, 1,..., n$,
and by the fact that $g=\sum^n_{i=0}a_if_i(c_j)=a_j$ for each $j$. 
Then $a_j=0$ for all $j$. 
Comes to the conclusion that {$f_0, f_1, ..., f_n$} is linearly independent.

First show that $f_i(c_j) = \delta_{ij}$, the kronecker delta. Then, upon assumption that $g = \alpha_0 f_0 + \cdots + \alpha_n f_n$ is the zero function, it will follow that $0 = g(c_j)=\sum_{i=0}^n \alpha_if_i(c_j) = \sum_{i=0}^n \alpha_i \delta_{ij} = \alpha_j$ for each $j$.
