Using the Levi-Civita alternating tensor and suffix notation to concisely write the vector product rule. I am reading through a section on vector calculus in an electromagnetism book and it has started to use suffix notation and the Levi-Civita alternating tensor in order to prove some identities.  Some of the identities I am familiar with and others I am not.  The notation is new to me as are the concept of tensors and I am struggling to apply both to do things which I can already do.

As an example it is stated in the book that it is much simpler and more concise to write the product rule as 

$$\left(\mathbf{A}\times\mathbf{B}\right)_i=\epsilon_{ijk}A_jB_k$$

and I know to work out the product rule I can use the determinant formula. What I am having trouble formulating in my head is how to actually read the above definition and get the correct expansion for the vector-product.  So to make sure I understood the notation I expanded it out on paper but ended up getting the wrong answer.  My interpretations and assumptions are as below:

$$\left(\mathbf{A}\times\mathbf{B}\right)_i = \epsilon_{ijk}A_jB_k$$
$$ = \epsilon_{ijk}A_yB_z\mathbf{i} + \epsilon_{jki}A_zB_x\mathbf{j} + \epsilon_{kij}A_xB_y\mathbf{k} + \epsilon_{jik}A_yB_z\mathbf{j} + \epsilon_{kji}A_zB_x\mathbf{k} + \epsilon_{ikj}A_xB_y\mathbf{i} $$

From here it can be seen that the first three terms are correct, but the last three are not.  I don't understand how the notation links back to the correct unit vector.  Here I have just take the first subscript letter in $\epsilon_{ijk}$ to also represent the relevant unit vector.  So I could really from getting the first three terms correct using the notation correctly write out the last three terms but that is just because I know what it should be.  What I do not understand is how the subscripts of the tensor link up to the subscripts of the two vectors.  I hope that is clear, if it isn't please leave a comment and I will try to remove anything confusing.
Actually, when you write $(A \times B)_i$, this is no longer a vector, but just one of its components. The vector itself is $$(A \times B) = ((A \times B)_1, (A \times B)_2, (A \times B)_3)$$
Having this in mind, the expression $$(A \times B)_i = \epsilon_{ijk}A_j B_k$$
means three equations, for $i$ ranges from $1$ to three. The equations are: $$(A \times B)_1 = \epsilon_{1jk}A_j B_k \\ (A \times B)_2 = \epsilon_{2jk}A_j B_k \\ (A \times B)_3 = \epsilon_{3jk}A_j B_k $$
For a better understanding of it, let's write everything for $(A \times B)_1$, say. I'll begin with the sum in $j$. So: $$\begin{align} (A \times B)_1 &amp;= \epsilon_{\color {red}{11}k}A_1B_k + \epsilon_{12k}A_2B_k + \epsilon_{13k}A_3B_k  \\ &amp;= \epsilon_{12k} A_2B_k + \epsilon_{13k} A_3 B_k
\end{align}$$
Notice that in the indices, repeats are zeros. Now, let's do the sum on $k$: $$\begin{align} (A \times B)_1 &amp;= \epsilon_{\color{red}{1}2\color{red}{1}}A_2 B_1 + \epsilon_{1\color {red}{22}}A_2B_2 + \epsilon_{123}A_2B_3 + \epsilon_{\color{red}{1}3\color{red}{1}}A_3B_1 + \epsilon_{132}A_3B_2 + \epsilon_{1\color{red}{33}}A_3B_3 \\ &amp;= \epsilon_{123}A_2B_3 + \epsilon_{132}A_3B_2 \\ &amp;= A_2B_3 - A_3B_2 \end{align}$$

You can do the same for the other ones, to get used to it. Also, I reccomend getting a few basic properties of the cross product, and prove them using only this notation. I once asked a question related to it, you might find it helpful.
Lost 1 is correct what you wrote is totally absurd....it is utter nonsense.
Recall :

$$\epsilon_{ijk} = -\epsilon_{ikj} = -\epsilon_{jki} = -\epsilon_{kji}.$$

We realize that product vanishes, i.e. it becomes zero....which I must say it's wrong for we don't have enough info to justify that.

So we rewrite the product as follows.

$$C_x = (\vec{A} \times \vec{B}) = \epsilon_{ijk} A_j B_k = \epsilon_{xyz} A_y B_z + \epsilon_{xzy} A_z B_y$$
I have ignored the other expressions since they lead to zero, i.e. 

$$\epsilon_{yzx} = \epsilon_{yxz} = \epsilon_{zyx} =\epsilon_{zxy} = 0,$$

since  in our expression we have $C_x$ and for the other components:

$$C_y = \epsilon_{yzx} - \epsilon_{yxz} \text{ and } C_z = \epsilon_{zxy} - \epsilon_{zyx}.$$