Question:
Suppose $p_0,p_1,..,p_m \in P(F)$ are such that each $p_j$ has degree $j$. Prove that $ p_0,p_1,..,p_m $ is a basis of $P_m(F)$, where $P_m(F)$ is the set of all polynomials with coefficients in $F$ and degree at most $m$.
Is this enough to prove the above statement? $\rightarrow$
Since each $p_i \in \{p_0,\dots,p_m\}$ is of degree $i$, that implies each $p_i$ is linearly independent.
Since $p_0...p_m$ is a linearly independent list of length $m+1$, it spans $P_m$, because the dimension of $P_m$ is of length $m+1$, and a linearly independent list of same dimension is a spanning list.
Your proof concerning the linear independenece of the list $p_0,p_1,\dots,p_m$ is incorrect .
What you need to do is to assume that for some $c_0,c_1,\dots c_m\in\mathbf{F}$ we have $q = \sum_{i=0}^{m}c_ip_i = 0$ and then use this to prove that $c_1 = c_2 = \dots = c_m = 0$ which you can do by repeatedly differentiating $q$ a total of $n+1$ times and each time infering that $c_i = 0$ after each differentiation.
Or alternatively you could do that by proving that $p_0$ alone is linearly independent and $p_j\not\in\operatorname{span}(p_0,p_1,\dots,p_{j-1}),\forall j\in\{1,2,\dots,m\}$.
Having proved linear independence then assuming that you know that $\dim\mathcal{P}_m(\mathbf{F}) = m+1$ you can infer that the given list is indeed a basis.
Hope you find the above useful. btw which text are you using?