Basis for $P_n$. Let $f \in \mathbb{R}[x]$ be a polynomial of degree $n$. Prove that the polynomials $f, f′, f′′,\ldots, f^{(n)}$ form a basis of the vector space $P_n$ of polynomials of degree $\leq n$.
 A: We need to make assumptions about the underlying field. This is because if that field has characteristic $p\ne 0$, and $f=x^p$, the derivatives are all $0$. (If you are working with polynomials with real coefficients, or complex coefficients, you need not worry about this.)
Hint: Our vector space has dimension $n+1$. We show the given bunch of polynomials is linearly independent. So suppose that 
$$a_0f+a_1f' +a_2f'' +\cdots +a_nf^{(n)}=0,\tag{$1$}$$
where the $a_i$ are scalars (numbers). We want to show that all the $a_i$ are $0$.  
First we show that $a_0= 0$. Suppose to the contrary that $a_0\ne 0$. The degree of $a_1f'+a_2f''+\cdots$ is $n-1$. So the degree of our sum on the left-hand side of $(1)$ is $n$, and therefore the sum cannot be the zero polynomial. 
So $a_0=0$. Essentially the same argument now shows that $a_1=0$, $a_2=0$, and so on. 
A: Hint: Write out a linear combination of $f, f', f'', \ldots$ and set it equal to zero.  Then look at the leading terms of the polynomials and try and prove that the coefficients must be zero.  This would prove that the set is linearly independent.  Then do you know what the dimension of $P_n$ is as a vector space?
