If $f(..., v_{i+1}, v_i,...) = −f(..., v_i, v_{i+1},...)$ prove that $f$ is alternating

Let f be a $k$-tensor on a vector space $V$. Prove that $f$ is alternating if and only if f changes sign whenever two successive arguments are interchanged: $f(..., v_{i+1}, v_i,...) = −f(..., v_i, v_{i+1},...)$

I have showed this $(\Rightarrow)$ direction. But I don't see how someone can show the other.

Let $σ$ be a permutation and $f(..., v_{i+1}, v_i,...) = −f(..., v_i, v_{i+1},...)$

$σf=f(v_{σ(1)},...,v_{σ(n)})$, for some $i\in \{1,2,...n\}$ it must be $σ(i)=1$ so with a finite interchanged of successive arguments I can bring $σ(i)=1$ to the $1st$ place meaning to transport $v_1$ to the $1st $ place, $v_2$ to the $2nd$ etc. so with a finite interchanged of successive arguments I can have $σf=(-1)^mf(v_1,...,v_n)$.

If the sum of those interchanged of successive arguments is an odd number ($m=$odd) it's obvious that $σf=f(v_1,...,v_n)$ and if it's even $σf=(-1)f(v_1,...,v_n)$.

In order to prove what I want, it's enough to show that if $σ$ is even, it corresponds to an even sum of interchanged successive arguments and if $σ$ is odd then it corresponds to an odd sum of interchanged successive arguments (or the opposite, I think would be the same)

I observe that if $σ$ is even (or same if it's odd), then that doesn't mean that the composition of transportations is successive $ \sigma=\begin{pmatrix}1&2&3&4&5\\ 3&4&5&2&1\end{pmatrix} \text{for instance} :σ = (2 3)(1 2)(2 4)(3 4)(1 5)$

Can someone give a hint, because I don't think I can prove this with my current knowledge.

If a permutation is even, there are lots of different ways of writing it as the composition of an even number of interchanges, but no ways of writing it as the composition of an odd number of interchanges. (and the reverse for odd permutations)

You can verify this by looking at the determinant of the corresponding permutation matrix. Any interchange has determinant -1, so any even permutation has determinant 1 and any odd permutation has determinant -1.