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: 
Write their trivial linear combination: $$a\_11 +a\_2(t-1)+a\_3(t^2-t)+\cdots +a\_n(t^{n-1}-t^{n-2})=0$$

so $$(a\_1-a\_2)1 + (a\_2-a\_3)t+(a\_3-a\_4)t^2+\cdots+(a\_{n-1}-a\_n)t^{n-2}+a\_nt^{n-1}=0 $$

is valid for all $t$ which means that this polynomial is actually zero polynomial, so $$a\_1-a\_2=a\_2-a\_3=...= a\_{n-1}-a\_{n}=a\_n=0$$

so all $a\_i$ are zero and thus a conclusion.
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.