Levi civita symbol identity with n dimension There is an identity $\displaystyle{\epsilon_{i_1...i_k i_{k+1}...i_n}\epsilon_{i_1...i_kj_{k+1}...j_n}  =k!\delta_{i_{k+1}...i_n }^{j_{k+1}...j_{n}}}$ in wikipedia. ( Some texts use different notation 
$\displaystyle{\epsilon_{{i_{k+1}}...{i_n}}\epsilon{_{{j_{k+1}...j_n}}}    }$.)
  https://en.wikipedia.org/wiki/Levi-Civita_symbol
I tried the expansion with the identity  $\displaystyle{\epsilon_{i_1...i_n}\epsilon_{j_1...j_n}  =\det(a_{lm})}$ where $a_{lm}=\delta_{i_l} \delta_{j_m}$. However, I can't find the $k!$ part.
When $k=1,n$, the proof is clear by properties of the determinant. 
How can I prove the general case for $k$?
 A: The following should be taken not so much as a proof but as a sketch thereof.
First, one shows that if the free indices are not permutations of one another, then both sides vanish trivially due to the definitions of $\epsilon$ and $\delta$. If they are permutations, then we can choose labels such that $j_{l}=l$ and $i_l =\sigma(l)$ for $l>k$ where $\sigma$ is some permutation on $n-k$ elements. One then argues that this can only modify both sides by the sign of the permutation $\sigma$, and therefore one can restrict to the case of $\sigma$ being the identity permutation.
Hence we need only consider the case of $i_{l}=j_{l}=l$ for $l>k$, in which the Kronecker delta on the RHS is simply 1 and the identity to be proven is $$\epsilon_{i_1 i_2 \cdots i_k k+1 \cdots n}\epsilon_{i_1 i_2 \cdots i_k k+1 \cdots n}=k!.$$
In order for the Levi-Civita symbols on the LHS to not vanish, we need $\{i_1,\cdots i_k\}$ to be some permutation $\sigma\in S_k$ of $\{1,2,\cdots k\}$. Hence we can simplify the Levi-Civita symbol to $\epsilon_{i_1 i_2 \cdots i_k}$ and the identity in this case becomes $\epsilon_{i_1 i_2 \cdots i_k}\epsilon_{i_1 i_2 \cdots i_k}=k!$. But this is equivalent to the case of $n=k$, which the OP noted to be obvious from the properties of the determinant. 
