I've been trying to prove the contracted epsilon identity ($ \varepsilon_{ijk}\varepsilon_{klm} = \dots$) with the help of this video. The proof writes the Levi-Civita symbol as a determinant, then multiplies the matrices and takes the determinant to get to the result.
I get the big picture of the proof, there's just one thing I don't really understand.
At some point, he gets the following value for the first row, first column of the matrix:
$$\delta_{i1}\delta_{1i}+\delta_{i2}\delta_{2i}+\delta_{i3}\delta_{3i}.$$
He says (around the 8:20 mark) that this equals $3$. Now, as far as I know, this can be written as:
$$\sum_{x=1}^3 \delta_{ix}\delta_{xi} = \delta_{ii}.$$
Why does this equal $3$? It would only be right if you were taking $\sum_{i=1}^3$ of the result.
The next part I understand, but I think he also made an error:
$$\delta_{j1}\delta_{1i}+\delta_{j2}\delta_{2i}+\delta_{j3}\delta_{3i}.$$
He says that equals $\delta_{jp}\delta_{pi}$. I say it equals $\sum_{p=1}^3\delta_{jp}\delta_{pi}$, though his final result is correct ($\delta_{ji}$)
I believe summation notation is being used. When a dummy index appears twice, it is being summed over all of its possible values.
So $\delta_{i1}\delta_{1i} + \delta_{i2}\delta_{2i} + \delta_{i3}\delta_{3i}$ really means
$$\sum_{i=1}^3(\delta_{i1}\delta_{1i} + \delta_{i2}\delta_{2i} + \delta_{i3}\delta_{3i}).$$
As $\delta_{xi} = \delta_{ix}$, $\delta_{ix}\delta_{xi} = \delta_{ix}^2 = \delta_{ix}$, the sum reduces to
\begin{align*} &\sum_{i=1}^3(\delta_{i1} + \delta_{i2} + \delta_{i3})\\ =& (\delta_{11} + \delta_{12} + \delta_{13}) + (\delta_{21} + \delta_{22} + \delta_{23}) + (\delta_{31} + \delta_{32} + \delta_{33})\\ =& (1 + 0 + 0) + (0 + 1 + 0) + (0 + 0 + 1)\\ =&\ 3. \end{align*}
As for the simplification of $\delta_{j1}\delta_{1i}+\delta_{j2}\delta_{2i}+\delta_{j3}\delta_{3i}$, your expression and his agree once you realise that he is using summation notation; his notation is shorthand for what you've written explicitly.