Problem: Any polynomial $P_n(x)$ can be written as $P_n(x)$=$\sum_{i=0}^n c_i \alpha_i(x) $ where $\alpha_i(x)$ is a polynomial of degree exactly $i$.
Attempt: Base case ($n=1$): $P_1(x)=c_0+c_1(x)$ and $\sum_{i=1}^1 c_i \alpha_i(x)= c_0+c_1(x)$.
Inductive hypothesis: Suppose this holds for $(n)$, and show that it holds for $(n+1)$. (Show that $P_{n+1}(x)=\sum_{i=0}^{n+1} c_i \alpha_i(x)$ using that $P_n(x)$=$\sum_{i=0}^n c_i \alpha_i(x)$)
Here's where I get confused; I don't quite understand the argument.
Hint $\ $ If $\,\deg P = n\,$ and $P$ has lead coef $\,p\,$ and $\,\alpha_n\,$ has lead coef $\,a\,$ then $\,P' = P - p/a\,\alpha_n$ has degree $< n$ since the lead terms cancel. So, by induction, $\, P' = c_1\alpha_1 + \cdots+ c_{n-1} \alpha_{n-1}\,$ thus $\,P = P' + p/a\,\alpha_n\,$ yields the sought form for $\,P.$
Remark $ $ We are essentially using the Polynomial Division Algorithm to divide $\,P\,$ by $\,\alpha_n\,\,$ so $\, P = q\, \alpha_n + r,\,$ but in the special case where the quotient $\,q\,$ is always a constant polynomial, since the divisor and dividend have equal degrees.