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If two indices of the n-dimensional Levi-Civita symbol are equal, exchanging them leaves the component unchanged. Antisymmetry of a multilinear map simultaneously negates it, so $ϵ=-ϵ$. Over the real numbers this implies $\boxed{ϵ=0}$.

Every nonzero component of the n-dimensional Levi-Civita symbol is $\pm 1$, so its square is one. There are exactly $n!$ permutations of the $n$ distinct indices; all remaining index lists have a repeat and contribute zero. Hence

If all $n$ indices are distinct and each lies in $\{1,\dots ,n\}$, their list is a permutation of $(1,\dots ,n)$. Sorting it takes some number $N$ of transpositions. The n-dimensional Levi-Civita symbol equals $(-1{)}^N$ because its sorted component is $1$. ThereforeThe parity of a permutation is independent of the chosen sequence of transpositions.