Linear Algebra Proof Question about vectors

Question: Assume that B = {V1,...,Vk} is a linearly independent set which spans R^n. Prove that every X which is part of R^n can be written as a unique linear combination of the vectors in B.

I know all the terminology ect but i'm not sure where to start on proofs.

C1V1 + ... + CkVk = 0 iff C = 0 (zero vector)
Let X be an element of R^n with
D1V1 + ... + DkVk = X for some D

Assume there exists some E such that
E1V1 + ... + EkVk = X
(D1V1 - E1V1) + ... + (DkVk - EkVk) = 0

Since B is linearly independent:
(D1V1 - E1V1) + ... + (DkVk - EkVk) = C1V1 + ... + CkVk
V1(D1 - E1) + ... + Vk(Dk - Ek) = C1V1 + ... + CkVk
so (D1 - E1) ... (Dk - Ek) must all be zero
so D = E
so X can be written as a unique linear combination of vectors in B

If you want to prove that something is unique, assume there are two different representations, and then show that leads to a contradiction. BTW, you have k vectors spanning R^n, so I'll assume k=n and just use n.

Assume A in R^n = sum of aiVi and also equals sum of biVi, and that ai does NOT equal bi for all i.
Then 0 = sum of (ai-bi)Vi, and not all ai-bi are 0.

First assume there is only 1 term where ai-bi isn't = 0, then aiVi = biVi, and so ai must = bi after all, a contradiction.

So assume there are more than 1 term where ai-bi = 0. Then there is some j where 0 =(aj - bj)Vj + [sum of some others]. so Vj = -1/(aj-bj) times [sum of some others]. So the elements of B are NOT linearly independent, another contradiction.

Since the assumption that the ai-bi are not all zero leads to a contradiction, it must be that ai=bi for all i, and hence that each element in R^n is a unique linear combination of the vectors in B.