Let $V$ be any vector space, (I'm not sure about the characteristic of the field) consider its tensor algebra $TV$. The two sided ideal is defined as the subspace $I \subset TV$ generated by all elements of the form $v\otimes v, v \in V$, i.e. by tensor multiplication and linear combination. Therefore its elements are of the form \begin{align} \sum_j x_j &\otimes v_j \otimes v_j \otimes y_j & v_j \in V,x_j,y_j &\in TV. \tag{*} \end{align} The exterior algebra is then defined as the quotient space $\Lambda V = TV/ I$, where one defines the projection $\pi:TV \to \Lambda V$ such that $v_1\otimes\dots\otimes v_k \mapsto [v_1\otimes\dots\otimes v_k]:= v_1\otimes\dots\otimes v_k + I$.
According to this answer, if I correctly understand it, any tensor $v_1\otimes\dots\otimes v_k$, where $v_1,\dots,v_k$ are linearly dependent, are in $I$ (the converse is of course clear.)
I have this way of proving it but I don't know how general it is.
My Attempt:
The problem can be reduced to proving that any decomposable with two equal factors (not necessarily adjacent) is in $I$, since a decomposable tensor which is a product of linearly dependent factors is a linear combination of tensors with two equal factors (not necessarily adjacent), i.e. $$ v_1,\dots,v_k \ \text{ are linearly dependent} \implies v_1\otimes\dots\otimes v_k = \sum_j \alpha_j (..\otimes w_j \otimes .. ..\otimes w_j \otimes..) $$ and $$ [..\otimes w_j \otimes .. ..\otimes w_j \otimes..] = [.. \otimes z\otimes z \otimes ..]. $$ I write only the inner part (surrounded by the equal factors) and without the class paranthises and with commas in place of the "$\otimes$". Let us establish some rules for transforming the representative element of a class by adding elements of the form $(*)$. Let us take class of the decomposable $a, b, c, d, a$. Then adding a decomposable $a,b,c,a,a$ gives $a,b,c,a+d,a$. Instead if we add $a,b,c,c,a$ then we get $a,b,c,d+c,a$. In total the allowed transformations on $a,b,c,d,a$ by adding elements of $I$ (of the form $(*)$) amount to scaling and adding any factor of $a,b,c,d$ to one its two neighbors. With this in mind, let's do this for our test tesnor \begin{align} \begin{matrix} a & b & c & d & a \\ a & b & c+d & d & a \\ a & b+(c+d) & c+d & d & a \\ a+(b+c+d) & b+c+d & c+d & d & a \\ a+b+c+d & b+c+d-(c+d) & c+d & d & a \\ a+b+c+d & b & c+d-d & d & a \\ a+b+c+d & b & c & d+a & a \\ a+b+c+d & b & c+(d+a) & d+a & a \\ a+b+c+d & b+(c+d+a) & c+d+a & d+a & a \\ a+b+c+d & a+b+c+d & c+d+a-(d+a) & d+a & a \\ a+b+c+d & a+b+c+d & c & d+a-a & a \\ a+b+c+d & a+b+c+d & c & d & a \\ \end{matrix} \end{align} so we achieved the first two factors on the left are equal, i.e. $$ [a\otimes b\otimes c\otimes d\otimes a] = [(a+b+c+d) \otimes (a+b+c+d) \otimes c \otimes d \otimes a]. $$ This should work for any $v_1 \otimes \dots \otimes v_k$ with $k > 5$.
Let $I_n$ be the subspace of $V^{\otimes n}$ generated by elementary tensors $v_1\otimes\cdots\otimes v_n$ in which two consecutive factors are equal, and let $f:V\times\cdots\times V\to V^{\otimes n}/I_n$ be the map such that $f(v_1,\dots,v_n)=[v_1\otimes\cdots\otimes v_n]$, with the brackets denoting classes modulo $I_n$. This is a multilinear function.
What you want to prove is that if $v_1,\dots,v_n$ are linerly dependent then $f(v_1,\dots,v_n)=0$.
To prove that, first show that $f$ vanishes whenever two of its arguments are equal, possibly by induction on the distance between its two equal arguments. Of course, by definition it vanishes whenever two consecutive arguments are equal. For example, we have that $$ f(x,y,z,x) = f(x+y,x+y,z,x) - f(x,x,z,x) -f(y,y,z,y)-f(y,x,z,x) $$ This reduces showing that $f(x,y,z,x)$, in which two arguments separated by $2$ places are equal, is zero, to showing that $f$ vanishes when two of its arguments separated by $1$ place are equal is zero.
Using that, suppose that one of the $v_i$ is a linear combination of the other vectors, replace it in $f(v_1,\dots,v_n)$ by that linear combination and use the linearity with respect to that argument.