$\frac{1}{k!}\sum_{h_1,...,h_k=1}^NA^{i_1,...,i_k}_{h_1,...,h_k}\big(A^{-1}\big)^{h_1,...,h_k}_{j_1,...,j_k}=\delta^{i_1,...,i_k}_{j_1,...,j_k}$ ACHTUNG
TO FOLLOWS I USE ENSTEIN CONVENTION TO WROTE DETERMINANTS!
Definition
If $A$ is an $n\times n$ matrix we define its $k\times k$ minotr to be the determinant of $k\times k$ submatrix of $A$, e.g.
$$
A^{i_1,...,i_k}_{j_1,...,j_k}:=\det\begin{bmatrix}A_{j_1}^{i_1}\ &\cdots&A_{j_k}^{i_1}\ \\\vdots&\ddots&\vdots\\A_{j_1}^{i_k}&\cdots&A_{j_k}^{i_k}\end{bmatrix}=\delta^{h_1,...,h_k}_{j_1,...,j_k}A^{i_1}_{h_1}...A^{i_k}_{h_k}=\delta^{i_1,...,i_k}_{h_1,...,h_k}A^{h_1}_{j_1}...A^{h_k}_{j_k}
$$
where the $\delta$-symbol is the generalised Kronecker Delta as here defined.
So I ask to prove that
$$
\frac{1}{k!}\sum_{h_1,...,h_k=1}^NA^{i_1,...,i_k}_{h_1,...,h_k}\big(A^{-1}\big)^{h_1,...,h_k}_{j_1,...,j_k}=\delta^{i_1,...,i_k}_{j_1,...,j_k}
$$
for any non-singular matrix. In particular using the above definition I calculate that
$$
\frac{1}{k!}\sum_{h_1,...,h_k=1}^NA^{i_1,...,i_k}_{h_1,...,h_k}\big(A^{-1}\big)^{h_1,...,h_k}_{j_1,...,j_k}=\\
\frac{1}{k!}\sum_{h_1,...,h_k=1}^N\Big(\delta^{l_1,...,l_k}_{h_1,...,h_k}A^{i_1}_{l_1}...A^{i_k}_{l_k}\Big)\Big(\delta^{h_1,...,h_k}_{m_1,...,m_k}\big(A^{-1}\big)^{m_1}_{j_1}...\big(A^{-1}\big)^{m_k}_{j_k}\Big)=\\
\frac{1}{k!}\sum_{h_1,...,h_k=1}^N\delta^{l_1,...,l_k}_{h_1,...,h_k}\delta^{h_1,...,h_k}_{m_1,...,m_k}A^{i_1}_{l_1}...A^{i_k}_{l_k}\big(A^{-1}\big)^{m_1}_{j_1}...\big(A^{-1}\big)^{m_k}_{j_k}=\\
\frac{1}{k!}\sum_{h_1,...,h_k=1}^N\delta^{l_1,...,l_k}_{m_1,...,m_k}A^{i_1}_{l_1}\big(A^{-1}\big)^{m_1}_{j_1}...A^{i_k}_{l_k}\big(A^{-1}\big)^{m_k}_{j_k}
$$
but unfortunately I do not be able to conclude anything. So could someone help me, please?
 A: The generalized delta is just an antisymmetrizer. You just need to use the commutation identity that you wrote at the end of the first displayed equation to slide the $A$'s through the antisymmetrizer.

Edit with a few more details:
I said the generalized delta
$$
\delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}=\sum_{\sigma\in S_k} {\rm sgn}(\sigma)
\delta^{i_{\sigma(1)}}_{j_1}\cdots \delta^{i_{\sigma(k)}}_{j_k}
$$
is an antisymmetrizer because if you contract it to a general tensor $T_{i_1,\ldots,i_k}$
and this way define a new tensor
$$
W_{i_1,\ldots,i_k}:= \delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}\ T_{j_1,\ldots,j_k}
$$
(I used Einstein's convention, but I don't care about upstairs/downstairs)
what you get is an antisymmetric tensor, i.e., $W_{i_1,\ldots,i_k}$ changes sign if any two indices are switched.
Now the commutation, or sliding of $A$'s through the antisymmetrizer is the identity
$$
\delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}A^{j_1}_{h_1}\cdots A^{j_k}_{h_k}=
A^{i_1}_{j_1}\cdots A^{i_k}_{j_k}\delta^{j_1,\ldots, j_k}_{h_1,\ldots, h_k}\ .
$$
As for the proof of the wanted identity, one can start as in the OP (a bad move that needs fixing)
$$
\frac{1}{k!} A^{i_1,\ldots,i_k}_{h_1,\ldots,h_k} (A^{-1})^{h_1,\ldots,h_k}_{j_1,\ldots,j_k}=
\frac{1}{k!} \left(
A^{i_1}_{p_1}\cdots A^{i_1}_{p_k}\delta^{p_1,\ldots, p_k}_{h_1,\ldots, h_k}\right)
\left(\delta^{h_1,\ldots, h_k}_{q_1,\ldots, q_k}(A^{-1})^{q_1}_{j_1}\cdots (A^{-1})^{q_k}_{j_k}\right)
$$
but the $A$'s and $A^{-1}$'s are on the wrong side of their attached generalized delta. So use the commutation twice to rewrite this as
$$
\frac{1}{k!}
\left(\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k}A^{p_1}_{h_1}\cdots A^{p_k}_{h_k}\right)
\left((A^{-1})^{h_1}_{q_1}\cdots (A^{-1})^{h_k}_{q_k}\delta^{q_1,\ldots, q_k}_{j_1,\ldots, j_k}\right)
$$
$$
=\frac{1}{k!}\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k}
(A^{p_1}_{h_1}(A^{-1})^{h_1}_{q_1})\cdots
(A^{p_k}_{h_k}(A^{-1})^{h_k}_{q_k})
\delta^{q_1,\ldots, q_k}_{j_1,\ldots, j_k}
$$
$$
=\frac{1}{k!}\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k}
\delta^{p_1}_{q_1}\cdots\delta^{p_k}_{q_k}
\delta^{q_1,\ldots, q_k}_{j_1,\ldots, j_k}
=\frac{1}{k!}\delta^{i_1,\ldots, i_k}_{p_1,\ldots, p_k}
\delta^{p_1,\ldots, p_k}_{j_1,\ldots, j_k}
=\delta^{i_1,\ldots, i_k}_{j_1,\ldots, j_k}\ .
$$

Edit with even more details:
For completeness, let me prove the identity about the last two contracted generalized deltas giving a single one.
By definition, and summing over permutation $\sigma,\tau$ in the symmetric group $S_k$,
$$
\delta^{i_1,\ldots,i_k}_{p_1,\ldots,p_k}\delta^{p_1,\ldots,p_k}_{j_1,\ldots,j_k}=
\sum_{\sigma,\tau}{\rm sgn}(\sigma){\rm sgn}(\tau)
\delta^{i_{\sigma(1)}}_{p_1}\cdots\delta^{i_{\sigma(k)}}_{p_k}
\ \delta^{p_{\tau(1)}}_{j_1}\cdots\delta^{p_{\tau(k)}}_{j_k}
$$
But, by permuting factors,
$$
\delta^{p_{\tau(1)}}_{j_1}\cdots\delta^{p_{\tau(k)}}_{j_k}=
\delta^{p_1}_{j_{\tau^{-1}(1)}}\cdots\delta^{p_k}_{j_{\tau^{-1}(k)}}\ .
$$
We insert this equality and do the summation over the $p$ indices and get
$$
\delta^{i_1,\ldots,i_k}_{p_1,\ldots,p_k}\delta^{p_1,\ldots,p_k}_{j_1,\ldots,j_k}=
\sum_{\sigma,\tau}{\rm sgn}(\sigma){\rm sgn}(\tau)
\delta^{i_{\sigma(1)}}_{j_{\tau^{-1}(1)}}\cdots\delta^{i_{\sigma(k)}}_{j_{\tau^{-1}(k)}}
$$
$$
=\sum_{\sigma,\tau}{\rm sgn}(\sigma){\rm sgn}(\tau)
\delta^{i_{\sigma(\tau(1))}}_{j_1}\cdots\delta^{i_{\sigma(\tau(k))}}_{j_k}
$$
by reordering the factors again. To finish, use
${\rm sgn}(\sigma){\rm sgn}(\tau)={\rm sgn}(\sigma\tau)$ and notice that each permutation $\rho=\sigma\tau$ appears exactly $k!$ times.
A: If $A$, $B$, $C$ are $(1,1)$ tensors such that $A\cdot B = C$, that is $\sum_h A^{i}_h B^h_j = C^i_j$. If we define
$A^{I}_J=A^{i_1, \ldots, i_k}_{j_1, \ldots, j_k} = A^{i_1}_{j_1}\cdots A^{i_k}_{j_k}$
we have $\sum_H A^I_H B^H_J = C^I_J$, which we can write
$(A)(B) = (AB)$ Consider now a special $(k,k)$ tensor, the generalized $\delta$, like @Abdelmalek Abdesselam: indicated,  and the tensor $[A] = \Delta \cdot (A)=(A)\cdot \Delta$ (last equality needs a bit of thinking, not true for any $(k,k)$ tenson). The crucial euqality is
$$\Delta \Delta = k! \Delta$$
Now we want to compare $[A]\cdot [B]$ and $[AB]$. So we write
$$[A]\cdot [B]= (A)\Delta \Delta (B)= k! \Delta (A)(B)= k! \Delta (AB) = k! [AB] $$
If instead of $\Delta$ we consider $P$ ( no signs), then we have permanents, and the corresponding Binet-Cauchy for them.
Note:
We have $(\Delta \cdot (A) )^I_J= \sum \epsilon(\sigma) (A)^{\sigma(I)}_J$ while  $(  (A) \cdot \Delta)^I_J= \sum \epsilon(\sigma) (A)^{I}_{\sigma(J)}$
Now the tensor $(A)$ is such that $(A)^{\sigma(I)}_{\sigma(J)} = (A)^I_J$ ( commutativity in some product), and moreover, $\epsilon(\sigma) = \epsilon(\sigma^{-1})$.
