Show $1, X-1, (X-1)^2, \ldots, (X-1)^{n-1}$ is a basis for $P_n(\mathbb R)$ (set of polynomials in $\mathbb R$ with degree $ \le n-1$). Show $1, X-1, (X-1)^2, \ldots, (X-1)^{n-1}$ is a basis for $P_n(\mathbb R)$ (set of polynomials in $\mathbb R$ with degree $ \le n-1$).
I have already proved $1,X,X^2,\ldots, X^{n-1}$ is a basis for $P_n(\mathbb R)$.
In proving $1, X-1, (X-1)^2, \ldots, (X-1)^{n-1}$ is a basis I must show the span equals $P_n(\mathbb R)$ and that these polynomials are independent.
Should I expand by binomial formula ? It gets quite messy.
 A: Let's prove the following more general fact: any family of polynomials $(p_0, \dots, p_{n-1})$ such that $\operatorname{deg}(p_k) = k$ is a basis for $P_{n}(\mathbb{R})$1.
To prove linear independence, use the property that $\operatorname{deg}(p+q) = \max\{\operatorname{deg}(p), \operatorname{deg}(q)\}$ if $\operatorname{deg}(p) \neq \operatorname{deg}(q)$.
Conclude by reasoning on the dimension of $P_{n}(\mathbb{R})$.

1 There is a name in French for such a family of polynomials (échelonné), there might be one in English too. 
A: Hint:  Try deduction on degree of polynomial $n$. Then you need only show that $X^{n-1}$ can be expressed by $1,X-1,\ldots,(X-1)^{n-1}$ and that $(X-1)^{n-1}$ is linear independent with $1,X-1,\ldots,(X-1)^{n-2}$.
A: Hint: you can also show that your collection of $n$ elements is a spanning set by showing inductively or otherwise that the span of these elements contains $1, X, X^2 \dots $ - a set which you know spans the space.
Note: If you already know that your space is a vector space (which is easy to show from first principles) then you can get away with linear independence or spanning set (you don't have to prove both), because every basis has the same number of elements.
A: Here is an abstract nonsense proof: Consider the shift $\sigma: \>x\to x-1$ on ${\mathbb R}$. Then the map $T:\ f\mapsto Tf:=f\circ\sigma$, which moves graphs of functions $f:\>{\mathbb R}\to{\mathbb R}$ one unit to the right, is linear, transforms polynomials into polynomials, does certainly not increase their degree, and has kernel $0$. It follows that $T':=T\restriction P_n({\mathbb R})$ is an automorphism of $P_n({\mathbb R})$; whence $T'$ maps any basis  of $P_n({\mathbb R})$ to a basis.
