A tensor of rank $n$ has components $T_{ij\cdots k}$ (with $n$ indices) with respect to each basis $\{\mathbf{e}_i\}$ or coordinate system $\{x_i\}$, and satisfies the following rule of change of basis: $$ T_{ij\cdots k}' = R_{ip}R_{jq}\cdots R_{kr}T_{pq\cdots r}. $$
Define the Levi-Civita symbol as: $$ \varepsilon_{ijk} = \begin{cases} +1 & ijk \text{ is even permutation}\\ -1 & ijk\text{ is odd permutation}\\ 0 & \text{otherwise (ie. repeated suffices)} \end{cases} $$
Show that $ \varepsilon_{ijk} $ is a rank 3 tensor.
I actually have a proof but I can't understand it! Can someone help me out?
$$ \varepsilon_{ijk}' = R_{ip}R_{jq}R_{kr}\varepsilon_{pqr} = (\det R)\varepsilon_{ijk} = \varepsilon_{ijk}, $$
This shows that $\varepsilon_{ijk}$ obeys the transformation law, so sure... but I don't follow what happened after the second equals sign
EDIT: Does this only hold for Cartesian coordinate systems, because then $R$ would be an orthogonal matrix with det 1 or -1?