Question: > 
> Prove that $1,t-1,t^2-t,\cdots,t^{n-1}-t^{n-2}$ a basis for any polynomial $P\_{n-1}$ represented by $a\_0+a\_1t+a\_2t^2+\cdots+a\_{n-1}t^{n-1}$
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$$
p(t)=a\_0+a\_1t+a\_2t^2+\cdots+a\_{n-1}t^{n-1}=
$$

Would be helpful I someone can give me a hint on how to rearrange the terms to show that the given set is a basis ?

Answer: 
Note that $S=\{1, t, t^2, \dots, t^{n-1}\}$ is a bais of $P\_{n-1}$.

Now consider the linear map $f\colon P\_{n-1}\to P\_{n-1}$ given by
$$
f(1) = 1,\ f(t) = t-1,\ f(t^2) = t^2-t,\ \dots,\ f(t^{n-1}) = t^{n-1} - t^{n-2}.
$$

The matrix of $f$ with respect to $S$ is obtained as
$$
\begin{pmatrix}
1 & -1 & 0 & \cdots & 0 \\
0 & 1 & -1 & \ddots & \vdots\\
0 & 0 & 1 & \ddots & 0 \\
\vdots & \vdots & \ddots & \ddots & -1 \\
0 & 0 & \cdots & 0 & 1
\end{pmatrix}.
$$
Since this matrix is invertible as an upper triangular matrix with non-zero diagonal entries, the linear map $f$ is an automorphism.

Hence, $f(S)$ is a basis as well.
The easiest way to show that some set is a basis for the set of polynomials, is to show that every element of the usual basis for the set of polynomials can be written as a sum of elements of your new basis.

In this case, it is therefore sufficient to show that $1, t, t^2, ..., t^{n-1}$ can be written as sums of elements of your new basis.

As $1 = 1, t = (t-1) + 1, t^2 = (t^2-t) + (t-1) + 1, ...$, this condition is satisfied, and therefore your new set is a basis for the set of polynomials.