Show that for each $n \in \mathbb{N}$, $\operatorname{span}\{p_1(x),\ldots,p_n(x)\} = \operatorname{span}\{1,x,x^2,\ldots,x^n\}$ Assume that, for each $n \in \mathbb{N}$, we have  $p_n(x)=\sum_{k=0}^{n-1}  x^k$ .
Show that for each $n \in \mathbb{N}$, $$\operatorname{span}\{p_1(x),\ldots,p_n(x)\} = \operatorname{span}\{1,x,x^2,\ldots,x^n\}$$
My approach is to say that $$\operatorname{span}\{p_1(x),\ldots,p_n(x)\} = \left(\sum_{k=1}^n \left(a_k \sum_{k=0}^{n-1} \right) \right) = \sum_{k=1}^na_kx^{k-1}$$
such that $n \in \mathbb{N}, (a_1,\ldots,a_n) \in \mathbb{R}, \sum_{k=0}^{n-1}  x^k \in \mathbb{R}[x] $ 
Is this correct?
 A: Your proof shows that any element spanned by $p_i(x)$ is also spanned by $x^i$. However, you also need to show the reverse to show that they're both equal. In my opinion, this is a bit hard, although certainly doable. However, I think it's easier to do this part by induction.
Base Case: Every element spanned by $p_1(x)=1$ is spanned by $x^0=1$. This is trivial to prove.
Induction Case: Assume $\text{span}\{p_1(x),p_2(x),...,p_n(x)\}=\text{span}\{1,x,x^2,...,x^{n-1}\}$. We want to prove that $\text{span}\{p_1(x),p_2(x),...,p_n(x),p_{n+1}(x)\}=\text{span}\{1,x,x^2,...,x^{n-1},x^n\}$. Again, you already proved that the former is a subset of the latter, so now we need to prove the reverse. From the latter, we have the following:
$$a_nx^n+a_{n-1}x^{n+1}+...+a_1$$
Now, comparing our assumption with what we want to prove, the new part here is the $x^n$ and the $p_{n+1}(x)$ elements. Now, $p_{n+1}(x)$ is the only element with a $x^n$ term in it, so if the element above is in the span of $p_i(x)$, the coefficient of $x^n$ must be the same as the coefficient of $p_{n+1}(x)$, so by separating $a_np_{n+1}(x)$ from the rest of the polynomial, we have:
$$(a_nx^n+a_nx^{n-1}+...+a_n)+(a_{n-1}-a_n)x^{n-1}+(a_{n-2}-a_n)x^{n-1}+...+(a_1-a_n)$$
$$a_np_{n+1}(x)+(a_{n-1}-a_n)x^{n-1}+(a_{n-2}-a_n)x^{n-1}+...+(a_1-a_n)$$
Now, the rest of the polynomial is of degree $x^{n-1}$, meaning it is spanned by $1,x,x^2,...x^{n-1}$. Thus, by our induction hypothesis, it is spanned by $p_1(x),p_2(x),...,p_n(x)$, so we have that this is equal to:
$$a_np_{n+1}(x)+\sum_{i=1}^n b_np_n(x)$$
Now, we have shown that any linear combination of $1,x,x^2,...,x^n$ can be written as a linear combination of $p_1(x),p_2(x),...,p_{n+1}(x)$ and thus $\text{span}\{1,x,x^2,...,x^{n-1},x^n\} \subseteq \text{span}\{p_1(x),p_2(x),...,p_n(x),p_{n+1}(x)\}$. Combine this with your proof and we have $\text{span}\{1,x,x^2,...,x^{n-1},x^n\} = \text{span}\{p_1(x),p_2(x),...,p_n(x),p_{n+1}(x)\}$, so we are done.

Now, as I hope you noticed, the above was very long. However, there's a way to do this without having to do this long proof: Change of basis.
You see, it's easy to prove that $1,x,x^2,...,x^{n-1}$ are linearly independent. Therefore, we can say that they are a basis for all polynomials of degree $n-1$ or less. This means that we can have a linear transformation that maps:
$$T(1)=(1,0,0,...,0)^t \in \Bbb{R}^n$$
$$T(x)=(0,1,0,0,,...,0)^t \in \Bbb{R}^n$$
$$T(x^2)=(0,0,1,0,0,...,0)^t \in \Bbb{R}^n$$
$$[...] \ T(x^{n-1})=(0,0,0,...,1)^t \in \Bbb{R}^n$$
Thus, we now have a vector representation of the polynomials. Now, we can represent the $p_i(x)$ as vectors in $\Bbb{R}^n$:
$$T(p_1(x))=T(1)=(1,0,0,...,0)^t$$
$$T(p_2(x))=T(1+x)=(1,1,0,0,...,0)^t$$
$$T(p_3(x))=T(1+x+x^2)=(1,1,1,0,0,...,0)^t$$
$$[...] \ T(p_n(x))=T\left(\sum_{i=0}^{n-1} x^i\right)=(1,1,1,...,1)^t$$
Now, we can use each of these mappings to create a matrix by using each vector as a column vector in our matrix:
$$\begin{matrix}1 & 1 & 1 & [...] & 1 \\ 0 & 1 & 1 & [...] & 1 \\ 0 & 0 & 1 & [...] & 1 \\ [...] \\ 0 & 0 & 0 & [...] & 1\end{matrix}$$
Clearly, this is an upper triangular matrix with non-zero diagonal entries, so it is non-singular. This means that the column vectors $p_i(x)$ are linearly independent. Thus, the set $\{p_1(x),p_2(x),...,p_n(x)\}$ is a span of $n$ linearly independent vectors, meaning their vector representations span $\Bbb{R}^n$ just like the $x^i$, so both sets have the same span.
Now, this way might seem like a very long method, too, but at least in my opinion, it's simpler to do this than to do the induction proof. (Although maybe that's just because I really like change of bases.)
A: I could not understand your approach very well (maybe there are some typos). But a simple proof is that 
(1) To show each member of $$\{p_1(x),\ldots,p_n(x)\}$$ can be obtained by a linear combination of the members of the other set, which is easy because the definition of $p_i(x)$ is based on the members of $\{1,x,x^2,\ldots,x^n\}$.
(2) To show that each member of $$\{1,x,x^2,\ldots,x^n\}$$ can be obtained as a linear combination of members of $\{p_1(x),\ldots,p_n(x)\}$, which is implied from the following:
$$x^k={p_k(x)-p_{k-1}(x)}$$.
