Deriving the Epsilon-Tensor (Levi-Civita Symbol) 
Given
$$\hat{e}_i \times \hat{e}_j=\sum_{k=1}^3\epsilon_{ijk}\hat{e}_k$$
and 
$\hat{e}_1, \hat{e}_2, \hat{e}_3$ are the unit vectors in a right
  handed cartesian coordinate system
Show that:
$$\epsilon_{ijk} = \begin{cases} 1,  & \text{if $\space i,j,k \space$
 is an even permutation of 1,2,3} \\[2ex]
 -1, & \text{if $\space i,j,k \space$ is an odd permutation of 1,2,3}\\[2ex] 0, & \text{if $\space i=j,j=k,k=i \space$} \end{cases}$$

What I have done so far:
I know that the cross product $\vec{a} \times \vec{b}$ yields a vector $\vec{c}$ that is orthogonal to the plane spanned by the vectors $\vec{a}$ and $\vec{b}$.
Writing out the sums:
$$\hat{e}_1 \times \hat{e}_2=\sum_{k=1}^3 \epsilon_{12k}\hat{e}_k=\epsilon_{121}\hat{e}_1+\epsilon_{122}\hat{e}_2+\epsilon_{123}\hat{e}_3 \space \space \space \space (1)$$
$$\hat{e}_2 \times \hat{e}_3=\sum_{k=1}^3 \epsilon_{23k}\hat{e}_k=\epsilon_{231}\hat{e}_1+\epsilon_{232}\hat{e}_2+\epsilon_{233}\hat{e}_3 \space \space \space \space (2)$$
$$\hat{e}_1 \times \hat{e}_3=\sum_{k=1}^3 \epsilon_{13k}\hat{e}_k=\epsilon_{131}\hat{e}_1+\epsilon_{132}\hat{e}_2+\epsilon_{133}\hat{e}_3 \space \space \space \space (3)$$
I know that equation $(1)$ needs to yield $\hat{e}_3$. 
$\implies\epsilon_{121}=\epsilon_{122}=0; \space \epsilon_{123}=1$
I know that equation $(2)$ needs to yield $\hat{e}_1$
$\implies \epsilon_{232}=\epsilon_{233}=0; \space \epsilon_{231}=1$
I know that equation $(3)$ needs to yield $-\hat{e}_2$
$\implies \epsilon_{131}=\epsilon_{133}=0; \space \epsilon_{132}=-1$
Is this a valid way of "proving" or showing the Levi-Civita symbol or am I doing something wrong? For some reason this doesn't seem right to me.
 A: You are correct. Your proof is totally valid. You can proceed in a similar fashion and prove the other values as well.However in your question, I don't think it will be $\epsilon_{ijk} >-1$ if $\space i,j,k \space$ is an odd permutation of $1,2,3$, rather it should be $\epsilon_{ijk}= -1$ if $\space i,j,k \space$ is an odd permutation of $1,2,3$
A: Your approach is valid, but you are not done yet, because there are $3^3=27$ values for $\epsilon_{ijk}$, and you have thus far calculated $9$.
$\newcommand{\vek}[1]{\boldsymbol{#1}}$
Another way to prove it is, to take your given formula and to do the dot product with $\vek{e}_k$
\begin{align}
(\vek{e}_i \times \vek{e}_j) \cdot \vek{e}_k 
= \sum_{l=1}^3 \epsilon_{ijl} ~\vek{e}_l \cdot \vek{e}_k
= \sum_{l=1}^3 \epsilon_{ijl} ~\delta_{lk}
= \epsilon_{ijk}
\end{align}
But $(\vek{e}_i \times \vek{e}_j) \cdot \vek{e}_k = \det(\vek{e}_i ~\vek{e}_j ~\vek{e}_k)$ so we get
$$
\det(\vek{e}_i ~\vek{e}_j ~\vek{e}_k) = \epsilon_{ijk}
$$
If you have $(ijk)=(123)$ then the determinant is just $1$. Because of the properties of the determinant, it stays $1$ for even permutations of $(\vek{e}_1 ~\vek{e}_2 ~\vek{e}_3)$ i.e. $(123)$, and changes the sign for odd ones. If at least two of $i,j,k$ are the same, then you have linear dependent vectors in your determinant, and hence it is $0$. So overall you get
$$
\epsilon_{ijk}
=
\det(\vek{e}_i ~\vek{e}_j ~\vek{e}_k)
= 
\begin{cases} 
1  & \text{if $\space (ijk) \space$ is an even permutation of (123) } \\
-1 & \text{if $\space (ijk) \space$ is an odd permutation of (123) } \\ 
0 & \text{if at least two of the $i,j,k$'s are the same} \end{cases}
$$
