Problem
Suppose $p_0, p_1, \dots, p_m \in \mathcal{P}(\mathbb{F})$ are such that each $p_j$ has degree $j$. Prove that $p_0, p_1, \dots, p_m$ is a basis of $\mathcal{P}_m(\mathbb{F})$
Solution
Proceed by induction on $m$.
BC: Let $m = 0$, then $p_0 = a_0z^0 = a_0$ is a basis for $\textrm{span}(a_0) = \textrm{span}(1) = \mathcal{P}_0(\mathbb{F})$.
IH: Suppose that for some $m = k \in \mathbb{N}$ we have $p_0, p_1, \dots, p_k$ is a basis for $\mathcal{P}_k(\mathbb{F})$ if $p_0, p_1, \dots, p_k \in \mathcal{P}(\mathbb{F})$ are such that each $p_j$ has degree $j$.
IS: If $p_0, p_1, \dots, p_k, p_{k+1} \in \mathcal{P}(\mathbb{F})$ are such that each $p_j$ has degree $j$. Then $p_{k+1}$ cannot be written in terms of the polynomials $p_0, p_1, \dots, p_k$, as the $z^{k+1}$ cannot be written as a linear combination of lower degree terms. Therefore by P2A11 and the IH, the list $p_0, p_1, \dots, p_k, p_{k+1}$ is therefore a basis for $\mathcal{P}_{k+1}(\mathbb{F})$. Therefore by PMI, it must hold for all $k \in \mathbb{N}_0$.