Prove that $\delta_{ij}\delta_{jk}=\delta_{ik}$ Firstly, I should mention that I have just started learning about tensors, the problem is that I need to understand why the result $$\fbox{$\delta_{ij}\delta_{jk}=\delta_{ik}$}\tag{1}$$ is true in order to be able to follow certain proofs given in my lecture course.
Before starting tensors I was taught that 
$$\delta_{ij} = \begin{cases}
0, & \mbox{if } i \ne j  \\
1, & \mbox{if } i=j \end{cases}\tag{2}$$
But now that I have been introduced to tensors I have been told that the Kronecker delta can actually be thought of as a tensor and that
$$\delta_{ij}=\begin{bmatrix}\delta _{11}&\delta _{12}&\delta _{13}\\\delta _{21}&\delta _{22}&\delta _{23}\\\delta _{31}&\delta _{32}&\delta _{33}\end{bmatrix}$$
for $i=1,2,3$ and $j=1,2,3$.
Using the definition in $\mathrm(2)$ then 
$$\delta_{ij}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I$$
I know that in $\mathrm(1)$ the Einstein summation convention is being used such that 
$$\delta_{ij}\delta_{jk}=\sum_{j=1}^{3}\delta_{ij}\delta_{jk}$$
But if I try to prove $(1)$ as follows
$$\delta_{ij}\delta_{jk}=\begin{bmatrix}\delta _{11}&\delta _{12}&\delta _{13}\\\delta _{21}&\delta _{22}&\delta _{23}\\\delta _{31}&\delta _{32}&\delta _{33}\end{bmatrix}\begin{bmatrix}\delta _{11}&\delta _{12}&\delta _{13}\\\delta _{21}&\delta _{22}&\delta _{23}\\\delta _{31}&\delta _{32}&\delta _{33}\end{bmatrix}$$
for $i=1,2,3$ and $j=1,2,3$ and $k=1,2,3$.
It doesn't bother me to multiply this whole thing out if it helps my understanding, but frankly, I don't see the point as 
$$\delta_{ij}\delta_{jk}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=I^2=I \ne \delta_{ik}$$
I have already watched this:
Why does $\delta_{ij}\delta_{jk}=\delta_{ik}\,$ An explanation to make sense!  which does not explain why $\delta_{ij}\delta_{jk}=\delta_{ik}$
I have also looked at a similar question on this site but the answer is too complicated for me to understand. 

So to summarize, is there a simple way of showing that $$\delta_{ij}\delta_{jk}=\delta_{ik}?$$
 A: Everyone struggles to gain intuition with this.  The way I came to realize it was to think about the terms in $\delta_{ij}\delta_{jk}$.  
$\delta_{ij}$ is zero unless $i=j$, and $\delta_{jk}$ is zero unless $j=k$.  So for any $j$, $\delta_{ij}\delta_{jk}$ (no summation) is $1$ if $i$, $j$, and $k$ are equal, and $0$ otherwise.
Now $i$, $j$, and $k$ are all being drawn from the same pool of cartesian indices.  So when we do sum over all $j$'s, exactly one of those $j$'s will agree with $i$, and exactly one will agree with $k$.  If $j$ agrees with $i$ at the same time it agrees with $k$ (that is, all three indices are equal), we will pick up a $1$.  For all the rest, we're summing up zeroes.  In other words, the sum is $1$ if $i=k$, and $0$ otherwise.  That tensor is $\delta_{ik}$.
A: $\delta_{ij}\delta_{jk}$ denotes a multiplication on the index j. $\delta_{ij}\delta_{jk}=\delta_{i1}\delta_{1k}+\delta_{i2}\delta_{2k}+\delta_{i3}\delta_{3k}$. If $i\neq 1$ and $k\neq 1$ for the first term, then the first term will be zero. Similarly  If $i\neq 2$ and $k\neq 2$ for the second term, then the second term will be zero. Therefore $\delta_{ij}\delta_{jk}=\delta_{11}\delta_{11}+\delta_{22}\delta_{22}+\delta_{33}\delta_{33}=\delta_{ik}$.
