Prove that if $(v_1,\ldots,v_n)$ spans $V$, then so does the list $(v_1-v_2,v_2-v_3,\ldots,v_{n-1}-v_n,v_n).$
Proof: Suppose that $V = \text{span } (v_1,\ldots,v_n).$ Then for any $v \in V$, there exist $a_1,\ldots,a_n \in \mathbb{F}$ such that
\begin{align} v &= a_1v_1+\cdots +a_nv_n \\ &= a_1v_1-a_1v_2 + a_1v_2+a_2v_2+a_3v_3+\cdots+ a_nv_n \\ &= a_1v_1-a_1v_2+a_1v_2-a_1v_3+a_2v_2-a_2v_3+a_1v_3+a_2v_3+a_3v_3+a_4v_4+\cdots+a_nv_n \\ &= a_1(v_1-v_2)+(a_1+a_2)(v_2-v_3)+(a_1+a_2+a_3)v_3+a_4v_4+\cdots+a_nv_n \\ &=\sum_{i=1}^{n-1} \left[\left(\sum_{k=1}^{i}a_k\right)(v_i-v_{i+1})\right] + a_nv_n \in \text{span}(v_1-v_2,v_2-v_3,\ldots,v_{n-1}-v_n,v_n). \end{align}
Now let $u \in \text{span}(v_1-v_2,v_2-v_3,\ldots,v_{n-1}-v_n,v_n).$ Then there exist $b_1,\ldots,b_n \in \mathbb{F}$ such that
\begin{align} u &= b_1(v_1-v_2)+ b_2(v_2-v_3)+\cdots +b_{n-1}(v_{n-1}-v_n) + b_nv_n \\ &= b_1v_1 + (b_2-b_1)v_2+(b_3-b_2)v_3+\cdots+(b_n-b_{n-1})v_n \in \text{span}(v_1,\ldots,v_n) \end{align} completing the proof. $\hspace{125mm} \Box$