Suppose that $v_1,v_2,v_3,v_4$ spans $V$. Prove that the list $v_1 - v_2, v_2 - v_3, v_3-v_4,v_4$ also spans $V$. Suppose that $v_1,v_2,v_3,v_4$ spans $V$. Prove that the list $v_1 - v_2, v_2 - v_3, v_3-v_4,v_4$ also spans $V$.
attempt: Suppose $v_1,v_2,v_3,v_4$ spans $V$, 
then let $v \in V$. 
So $v = a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4$ for some $a_1,a_2,a_3,a_4 \in F$. 
Then we will show $v \in span(v_1 - v_2, v_2 - v_3, v_3-v_4,v_4) $.
So $v = c_1(v_1 -v_2) + c_2(v_2-v_3) + c_3(v_3 - v_4) + c_4v_4$ for some $c_1,..,c_4 \in F.$.
Then setting the two equations equal we have $a_1v_1 + a_2v_2 + a_3v_3 + a_4v_4 = c_1(v_1 -v_2) + c_2(v_2-v_3) + c_3(v_3 - v_4) + c_4v_4$.
I am stuck and don't really know how to conclude. Can someone please help  me?
Thank you

Set $c_1=a_1$. This is going to produce the same coefficient accompanying $v_1$. Then we have to mend $v_2$. For that, set $c_2=a_1+a_2$. Then we have to mend $v_3$, and so on... The idea is that each vector of $v_1 - v_2, v_2 - v_3, v_3-v_4,v_4$ is going to mend the previous one, until the last one, $v_4$, fixes itself. It is easily generalizable for n vectors , and this is a way( my way) to prove that if a vector space has a basis, then it has infinite of them.
