I want to prove that $i_v(T\wedge S)=(i_vT)\wedge S+(-1)^kT\wedge(i_vS)$
where $T\in \lambda^k V^*$, and $S\in \lambda^lV^*$.
For that I have two clues.
First I saw the advice to begin with the particular case $T=\xi_1\wedge\ldots\xi_l$ and $S=\sigma_1\wedge\ldots\wedge \sigma_l$. In that particular case, I can use the formula
$(\xi_1\wedge\ldots \wedge \xi_k)(v_1,\ldots,v_k)=\det\big( \xi_i(v_j)\big)$
and maybe the invariance of the determinant by permutation of columns ?
The second clue I have is to divide the symmetric group $S_{k+l}$ in two parts :
$A=\{ \sigma\in S_{k+l}\,s.t. \sigma(1)\leq k \}$
$B=\{ \sigma\in S_{k+l}\,s.t. \sigma(1)> k \}$
When developing $i_v(T\wedge S)$, hopefully the part of the sum in $A$ will provide the term $(i_vT)\wedge S$ because the $v$ will remain in the "$T$ part".
Anyway.
QUESTION : How to prove $i_v(T\wedge S)=(i_vT)\wedge S+(-1)^kT\wedge(i_vS)$ ?