Let $V_n$ denote the vector space of polynomials of degree at most $n$ over a field $F$, and suppose $\{f_0, f_1, \cdots, f_{n-1}, f_n\}$ is a set of polynomials such that for all $j$ between $0$ and $n$, we have that $f_j$ has degree $j$. Show that the $\{f_j\}$ form a basis for $V_n$.
Let $B = \{f_0, f_1, \cdots, f_{n-1}, f_n\}$. We want to show that $B$ is linearly independent and spans $V_n$.
For linear independence, suppose $$\sum_{i=0}^{n}a_if_i = \sum_{i=0}^{n}a_i(\sum_{j=0}^{i}a_{ij}x^j) = \sum_{i=0}^{n}(\sum_{j=0}^{i}a_ia_{ij}x^j) = \sum_{j=0}^{n}(\sum_{i=j}^{n}a_ia_{ij})x^j = 0$$ Then, $\sum_{i=j}^{n}a_ia_{ij} = 0$ for all $j$. Specifically, for $j=n$, $a_na_{nn} = 0$. $a_{nn}$ cannot be 0 since $x^j$ for $j=n$ must be of degree $n$, and so $a_n = 0$. Thus, we can rewrite as $$\sum_{i=0}^{n}a_if_i = \sum_{i=0}^{n-1}a_if_i +a_{n}f_n = \sum_{i=0}^{n-1}a_if_i + 0 = \sum_{j=0}^{n-1}(\sum_{i=j}^{n-1}a_ia_{ij})x^j = 0$$ We do the same process for $j = n-1$ to conclude that $a_{n-1}=0$ until $j=0$. Thus, $a_i=0$ for all $i$, and so $B$ is linear independent.
For spanning …
EDIT: Attached below is answer key's proof, though I'm a bit confused by it.
Best Answer
Since $\dim V_n=n+1$ and since $B$ has $n+1$ linearly independent vectors, $B$ spans $V_n$.
This has nothing to do with your specific situation. If $V$ is a $n$-dimensional vector space over some field $F$ and if $S\subset V$ is a set with $n$ elements, then the following assertions are equivalent: