Weinberg's proof that Levi-Civita Symbol is a tensor In Gravitation and Cosmology, S.Weinberg states the following:
$$\Lambda_{\epsilon}^{\alpha}\Lambda_{\zeta}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda} \propto \epsilon^{\alpha \beta \gamma \delta} $$
and the argument is that the left-hand side must be odd under a single permutation of the indices. I don't see why this is true. Say I interchange $\alpha\leftrightarrow \beta$:
$$\Lambda_{\epsilon}^{\beta}\Lambda_{\zeta}^{\alpha}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda} $$
I don't see why the above expression must satisfy
$$\Lambda_{\epsilon}^{\beta}\Lambda_{\zeta}^{\alpha}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda}=-\Lambda_{\epsilon}^{\alpha}\Lambda_{\zeta}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda} $$
Any hint will be appreciated.
 A: First of all, by interchanging $\Lambda_{\epsilon}^{\beta}$ with $\Lambda_{\zeta}^{\alpha}$ nothing changes: 
$$\Lambda_{\epsilon}^{\beta}\Lambda_{\zeta}^{\alpha}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda}=\Lambda_{\zeta}^{\alpha}\Lambda_{\epsilon}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda}$$
Now switch $\epsilon$ with $\zeta$ only in the Levi-Civita symbol:
$$\Lambda_{\zeta}^{\alpha}\Lambda_{\epsilon}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda} = -\Lambda_{\zeta}^{\alpha}\Lambda_{\epsilon}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\zeta \epsilon \kappa \lambda}$$
and rename all the $\epsilon$ with $\zeta$ and all the $\zeta$ with $\epsilon$ (again nothing changes):
$$-\Lambda_{\zeta}^{\alpha}\Lambda_{\epsilon}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\zeta \epsilon \kappa \lambda}\to-\Lambda_{\epsilon}^{\alpha}\Lambda_{\zeta}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta  \kappa \lambda}$$
So from (1),(2) and (3) you get
$$\Lambda_{\epsilon}^{\beta}\Lambda_{\zeta}^{\alpha}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta \kappa \lambda} = -\Lambda_{\epsilon}^{\alpha}\Lambda_{\zeta}^{\beta}\Lambda_{\kappa}^{\gamma}\Lambda_{\lambda}^{\delta}\epsilon^{\epsilon \zeta  \kappa \lambda}$$
A: Use the anti-symmetry of $\epsilon$ to switch the indices $\epsilon$ and $\zeta$. Then relabel your dummy indices.
EDIT: Allow me to expand on this answer on Bebop's behalf. 
You have two steps. In the first step, you use the fact that $\epsilon$ is anti-symmetric. That means you'll get a minus-sign if you exchange $\zeta$ and $\epsilon$. In the second step, you use the fact that you're ultimately summing over those indices. That means that $\zeta$ and $\epsilon$ are so-called dummy variables. Their names don't mean anything outside the sum, and so we can rename them any way we want. The idea here is basically that $\sum_n a_n = \sum_m a_m$. This means that you can just as well flip $\zeta$ and $\epsilon$ again, but this time "everywhere" in the expression (in $\epsilon$ and $\Lambda$), and this time without a sign change.
This technique (flipping first using the symmetry or antisymmetry and then a second time using the fact that we just have dummy variables) comes up quite a lot in physics.
