Prove that if $(v_1,\ldots,v_n)$ spans $V$, then so does the list $(v_1-v_2,v_2-v_3,\ldots,v_{n-1}-v_n,v_n).$

Just for fun you can prove this more general result.

Let $V$ be an $n-$dimensional vector space, and let $\vec{v}\in V$. Assume that $\left( \vec{\beta}_1,\ldots ,\vec{\beta}_n\right)$ is a basis for $V$. If $\vec{v}=c_1\vec{\beta}_1 +\cdots+ c_n \vec{\beta}_n$, then for $c_j \ne 0$, we can replace one of the $\beta_j$'s with $v$ so that $\left( \vec{\beta}_1,\ldots , \vec{\beta}_{j-1},\vec{v},\vec{\beta}_{j+1},\ldots, \vec{\beta}_{n}\right)$ is a basis for $V$.

So you can use it $n$ times for each element in your list.