You got confused with the notation. At one moment, the hint is using the subscript on fr to determine the function, i.e. fr = (1, r, r2, ...). Next, they are using the subscripts to index, so when they talk about f1, ..., fn, they just mean any n functions (from T), i.e. not that fn = (1, n, n2, ...). To be more clear, let's instead think about
f_{r_1},\, f_{r_2},\, \dots ,\, f_{r_n}
If these are linearly dependent, then there exists a set of reals, {a1, ..., an} not all zero such that:
\sum _{i = 1} ^n a_if_{r_i} = 0
Thus, for ANY vector v in V, we get:
\sum _{i = 1} ^n a_if_{r_i}(v) = 0(v) = 0
In particular, this holds (simultaneously) for v0, v1, ..., vn-1, so we get:
\sum _{i = 1} ^n a_if_{r_i}(v_0) = \sum _{i=1} ^n a_i = 0
\sum _{i = 1} ^n a_if_{r_i}(v_0) = \sum _{i=1} ^n a_ir_i = 0
.
.
.
\sum _{i = 1} ^n a_if_{r_i}(v_0) = \sum _{i=1} ^n a_ir_i^{n-1} = 0
So:
\left(<br />
\begin{array}{cccc}<br />
1 & 1 & \dots & 1\\<br />
r_1 & r_2 & \dots & r_n \\<br />
\vdots & \vdots & \ddots & \vdots \\<br />
r_1^{n-1} & r_2^{n-1} & \dots & r_n^{n-1}<br />
\end{array}<br />
\right)<br />
\left(<br />
\begin{array}{c}<br />
a_1\\<br />
a_2\\<br />
\vdots \\<br />
a_n<br />
\end{array}<br />
\right)<br />
=<br />
\left(<br />
\begin{array}{c}<br />
0\\<br />
0\\<br />
\vdots \\<br />
0<br />
\end{array}<br />
\right)
You should be able to prove that the ai are all zero from here. Prove it by contradiction, and think about how many roots a k-degree polynomial can have. The fact that T is linearly independent then follows immediately. I don't think you need to show that T spans your space, because showing that T is a basis is overkill, is it not? If T is uncountable and linearly independent, then surely any (not the) basis for your space is uncountable.