If $f\in A_k(V)$ and $g\in A_l(V)$ show $i_v(f\wedge g)=i_vf\wedge g+(-1)^kf\wedge i_vg$ - I've got the gist, not sure how to write If $f\in A_k(V)$ and $g\in A_l(V)$ show $i_v(f\wedge g)=i_vf\wedge g+(-1)^kf\wedge i_vg$
With $A_k(V)$ being the vector space of alternating k-tensors.
for $f\in A_k(V)$ for some $v\in V$ we define $i_vf\in V_{k-1}(V)$ by $i_vf(v_1,v_2,...,v_{k-1})=f(v,v_1,v_2,...,v_{k-1})$
My "answer"
This is similar to combinatorics, we all know Pascal's relation: $(n+1)Cr=nC(r-1)+nCr$ the proof is simple, given a set of $\{a_1,...,a_n,b\}$ (b is the $n+1$th thing) and tasked with choosing r things from it
We may have b then $r-1$ things from the remaining $n$, or!
We do not have b then $r$ things from the remaining $n$
I think that's at play here.
why:
Well $$f\wedge g(a_1,...,a_{k+l})=\sum_{\sigma\in S_{k+l}}\text{sign}(\sigma)f(a_{\sigma_1},...,a_{\sigma_k})g(a_{\sigma_{k+1}},...,a_{\sigma_{k+l}})
$$
So $$i_v(f\wedge g)(a_1,...,a_{k+l-1})=f\wedge g(v,a_1,...,a_{k+l-1})$$
For a permutation of $\{v,a_1,...,a_{k+l-1}\}$ either v is an argument to f or it is not. If the $v$ is an argument to $f$ then we're somewhere with the $i_vf\wedge g$ part. Otherwise the $v$ is one of g's argument.
This is what I mean by "gist" of argument, I'm struggling to form this into the expression I want as a result.
 A: Both expressions $\iota_v(f\wedge g)$ and $\iota_vf\wedge g+(-1)^kf\wedge\iota_vg$ are linear in $f$ and $g$ so it suffices to prove the claim on basis elements. That is, assume $f=\alpha^1\wedge\dots\wedge\alpha^k$ and $g=\alpha^{k+1}\wedge\dots\wedge\alpha^{k+l}$ where each of $\alpha^1,\dots,\alpha^{k+l}$ are all $1$-forms. One way to prove the claim is to use the fact that $$\iota_v(\alpha^1\wedge\dots\wedge\alpha^n)=\sum_{i=1}^k(-1)^{i-1}\alpha^i(v)\alpha^1\wedge\dots\wedge\alpha^{i-1}\wedge\alpha^{i+1}\wedge\dots\wedge\alpha^n$$ You can prove this fact by recalling that $$\alpha^1\wedge\dots\wedge\alpha^n(v,v_2,\dots,v_n)=\text{det}\left[\begin{array}{c c c c}\alpha^1(v)&\alpha^1(v_2)&\dots&\alpha^1(v_n)\\\alpha^2(v)&\alpha^2(v_2)&\dots&\alpha^2(v_n)\\\vdots&\vdots&\ddots&\vdots\\\alpha^n(v)&\alpha^n(v_2)&\dots&\alpha^n(v_n)\end{array}\right]$$ and performing cofactor expansion along the first column. We have by definition, $\iota_v(f\wedge g)=\iota_v(\alpha^1\wedge\dots\wedge\alpha^{k+l}),$ while by the above this is equal to $$\sum_{i=1}^{k+l}(-1)^{i-1}\alpha^i(v)\alpha^1\wedge\dots\wedge\alpha^{i-1}\wedge\alpha^{i+1}\wedge\dots\wedge\alpha^{k+l}$$which is $$\begin{array}{c}\left(\sum_{i=1}^k(-1)^{i-1}\alpha^i(v)\alpha^1\wedge\dots\wedge\alpha^{i-1}\wedge\alpha^{i+1}\wedge\dots\wedge\alpha^{k}\right)\wedge\alpha^{k+1}\wedge\dots\wedge\alpha^{k+l}\\ + \ \ \ \alpha^{1}\wedge\dots\wedge\alpha^{k}\left(\sum_{i=1}^{l}(-1)^{k+i-1}\alpha^{k+i}(v)\alpha^{k+1}\wedge\dots\wedge\alpha^{k+i-1}\wedge\alpha^{k+i+1}\wedge\dots\wedge\alpha^{k+l}\right)\end{array}$$ which is $$\begin{array}{c}\left(\sum_{i=1}^k(-1)^{i-1}\alpha^i(v)\alpha^1\wedge\dots\wedge\alpha^{i-1}\wedge\alpha^{i+1}\wedge\dots\wedge\alpha^{k}\right)\wedge\alpha^{k+1}\wedge\dots\wedge\alpha^{k+l}\\ +\ \ \ (-1)^k\alpha^{1}\wedge\dots\wedge\alpha^{k}\left(\sum_{i=1}^{l}(-1)^{i-1}\alpha^{k+i}(v)\alpha^{k+1}\wedge\dots\wedge\alpha^{k+i-1}\wedge\alpha^{k+i+1}\wedge\dots\wedge\alpha^{k+l}\right)\end{array}$$ which is 
$$\iota_vf\wedge g+(-1)^kf\wedge\iota_vg$$
