Proof the Levi-Civita symbol is a tensor 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?

I met this problem today and this is what I am trying:
$$
\epsilon_{ijk}=det(e_i\ e_j \ e_k)
$$
Let A be an orthogonal transformation, then:
$$
\begin{aligned}
\epsilon'_{ijk}&\equiv\epsilon_{lmn}=det( e_l \ e_m \ e_n)\\
&=det(Ae_i\ Ae_j \ Ae_k)\\
&=(det(A))^3 det(e_i e_j e_k)\\
&=det(A)\cdot \epsilon_{ijk}\\
&=\pm \epsilon_{ijk}
\end{aligned}
$$
I think this can show that the Levi-Civita is a tensor (or pseudotensor).
