Answer
See below
Work Step by Step
Given every vector $v \in V$ can be written uniquely as a linear combination of the vectors $\{v_1,v_2,...,v_n\}$ and $\{v_1,v_2,...,v_n\}$ as linearly independent set we have $a_1,a_2,...a_n$
Since $\lambda_1v_1+\lambda_2v_2+...\lambda_nv_n=0\\
\rightarrow v=a_1v_1+a_2v_2+...a_nv_n+\lambda_1v_1+....\lambda_n v_n\\
\rightarrow v=(a_1+\lambda_1)v_1+...$
There always at least some $\lambda_i \ne 0 \rightarrow a_i+\lambda_1\ne a_i$
Since $dim (V)=n$ number of vectors, hence $v_1,v_2,...v_n$ is a base for $V$
