$\sum_{i_{K+1},...,i_N=1}^N ε^{i_1,...,i_K,i_{K+1},...,i_N}δ^{j_{K+1},...j_N}_{i_{K+1},...,i_N}=(N-K)!ε^{i_1,...,i_K,j_{K+1},...,j_N}$ What shown below is a reference from Introduction to vectors and tensors by Ray M. Bowen and C.C Wang. Precisely it is de definition of the generalised Kroneker delta thus if you know it you can even not read what is written in the image and so you can directly read what I ask to prove.

So using this definition I ask to prove that

*

*$$
\sum_{i_{K+1},...,i_N=1}^N ε^{i_1,...,i_K,i_{K+1},...,i_N}δ^{j_{K+1},...j_N}_{i_{K+1},...,i_N}=(N-K)!ε^{i_1,...,i_K,j_{K+1},...,j_N}
$$
So could someone help me, please?

So we let's start to analyse the summation to the left hand of the equality $(1)$. So if
$$
ε^{i_1,...,i_K,i_{K+1},...,i_N}δ^{j_{K+1},...j_N}_{i_{K+1},...,i_N}\neq 0
$$
then
$$ε^{i_1,...,i_K,i_{K+1},...,i_N}\neq 0\,\,\,\text{and}\,\,\,δ^{j_{K+1},...j_N}_{i_{K+1},...,i_N}\neq 0
$$
and so
$$
\{i_{K+1},...,i_N\}\subseteq\{1,...,N\}\setminus\{i_1,...,i_K\}
$$
and thus $I:=\{i_1,...,i_K\}$ and $J:=\{j_{K+1},...,j_N\}$ determine a partition of $\{1,...,N\}$ because $δ^{j_{K+1},...j_N}_{i_{K+1},...,i_N}\neq 0$ and this implies $\{j_{K+1},...j_N\}=\{i_{K+1},...,i_N\}$. So if the sets $I$ and $J$ do not determine a partition of $\{1,...,N\}$ then clearly any term of the summation to the left hand is equal to zero and the same is for the right hand so that the equality trivially holds. So now we suppose that the sets $I$ and $J$ determine a partition of $\{1,...,N\}$ and we analyse which terms to the summation are not zero. So by the previous observation we know that the only indices $i_{K+1},...,i_N$ such that the corresponding term of the summation to the left hand of the equation $(1)$ is not zero are only that whose indices determine a permutation of $J$ so that we have $(N-K)!$ not zero terms. So if we now observe that
$$
ε^{i_1,...,i_K,i_{K+1},...,i_N}δ^{j_{K+1},...j_N}_{i_{K+1},...,i_N}=
\\ε^{i_1,...,i_K,i_{K+1},...,i_N}δ^{i_1,...,i_K,j_{K+1},...j_N}_{i_1,...,i_K,i_{K+1},...,i_N}=
\\
\\ε^{i_1,...,i_K,i_{K+1},...,i_N}ε^{i_1,...,i_K,j_{K+1},...j_N}ε_{i_1,...,i_K,i_{K+1},...,i_N}=
\\ε^{i_1,...,i_K,i_{K+1},...,i_N}ε_{i_1,...,i_K,i_{K+1},...,i_N}ε^{i_1,...,i_K,j_{K+1},...j_N}=
\\δ^{i_1,...,i_K,i_{K+1},...,i_N}_{i_1,...,i_K,i_{K+1},...,i_N}ε^{i_1,...,i_K,j_{K+1},...j_N}=
\\ε^{i_1,...,i_K,j_{K+1},...j_N}
$$
and thus the equation follows immediately.
