Answer
See below
Work Step by Step
Given $V_n=P_n(R)$ and $S=\{p_1,p_2,...,p_k\}$
Assume without loss of generality that the polynomials are ordered in descending degree:
$deg(p_1) \lt deg (p_2) \lt deg (p_3)\lt...\lt deg(p_k)$
Since $deg(c_1p_1+c_2p_2+...+c_{k-1}p_{k-1})\lt deg (p_k)$
and $deg(p_i)=n\\
deg(p_j)=m\\
\rightarrow p_k=a_0+a_1x+a_2x^2+...+a_mx^m$
then $a_m=0\\
c_k=0$
Obtain: $c_1p_1+c_2p_2+...+c_kp_k=0\\
\rightarrow c_1p_1+c_2p_2+...+c_{k-1}p_{k-1}=0$
Since $deg (P_{k-1}) \gt deg (\sum c_kp_k)$ we have $c_{k-1}=0$
If we repeat this $k-3$ times we get $c_{k-2}=c_{k-3}=...=c_1=0$
Hence, $p_i$ are linearly independents.
