Let $R$ be a commutative ring with unity , consider the polynomial ring $R[x]$ , let $\mathcal P_n:=\{f \in R[x] : f=0$ or $\deg f \le n\}$ , so $\mathcal P_n$ is a finitely generated module over $R$ . Let $a \in R$ be a unit(invertible) and $p_1(x),...,p_n(x) \in R[x]$ be polynomials with invertible leading coefficients such that $\deg p_i(x)=i , \forall i=1,...,n$ ; then does the set $\{a,p_1(x),...,p_n(x)\}$ generate $\mathcal P_n$ ?
[ If the answer to the question is yes , then in this way , I can prove the division algorithm for $R[x]$ namely , let $f(x) , g(x) \in R[x]$ be polynomials with $g(x)$ non-zero and leading co efficient of $g(x)$ be unit ; by passing the trivial cases like $f(x)=0$ or $\deg f < \deg g$ or $\deg f = \deg g = 0$; let $\deg f \ge \deg g >0$ , let $\deg f=n , \deg g =m $ , then consider the set of $n+1$ elements $\{1,x,...,x^{m-1}, g,xg,...,x^{n-m}g\}$ , if the answer to the question is yes , then this set generates the $R$-module $\mathcal P_n$ , then there exist $a_0,...,a_{m-1} , a_m,...,a_n \in R$ such that $f(x)=a_0+a_1x+...+a_{m-1}x^{m-1}+(a_m+a_{m+1}x+...+a_nx^{n-m})g(x)=r(x)+q(x)g(x)$ ,
where $r(x):=a_0+a_1x+...+a_{m-1}x^{m-1} , q(x):=a_m+a_{m+1}x+...+a_nx^{n-m}$ with
$\deg r(x) \le m-1 < m = \deg g $ , thus establishing the division algorithm . ]
Letting $p_0=a_0=a$, your hypothesis means that $$ \pmatrix{ p_0 \\ p_1 \\ \vdots \\ p_n } = \pmatrix { a_0 & 0 & 0 & 0 \\ * & a_1 & 0 & 0 \\ * & * & \ddots& 0 \\ * & * & * &a_n } \pmatrix{ 1 \\ x \\ \vdots \\ x^n } $$ The matrix is lower triangular with invertible elements in the diagonal. Therefore, the matrix is invertible and you can write $1,x,\ldots,x^n$ as a linear combination of $p_0, p_1, \ldots, p_n$.