Let $$N_{ij} = \delta_{ij} - \epsilon_{ijk}n_k + n_in_j$$ and $$M_{ij}= \delta_{ij} + \epsilon_{ijk}n_k$$ where $n_i$ is the cartesian coordinates of the unit vector $\textbf{n}$.
Show that $$N_{ij}M_{jk}=2\delta_{ik}$$ Below is my working so far:
$N_{ij}M_{jk}=(\delta_{ij} - \epsilon_{ijl}n_l + n_in_j)(\delta_{jk} + \epsilon_{jkl}n_l)$ $\hspace{2em}$ [introducing new variable 'l']
=$\delta_{ik}-\epsilon_{ijl}\epsilon_{jkl}n_ln_l+\delta_{jk}n_in_j+\epsilon_{jkl}n_ln_in_j$
=$\delta_{ik}+2\delta_{ik}+\delta_{jk}n_in_j+\epsilon_{jkl}n_ln_in_j$
This is where I got stuck, since I am not familiar with the summation convention- especially with multiple terms.
NB - What should one do if an index appears more than twice in one term for Einstein convention?
Thank you in advance.
The $l$ only appears more than twice because you decide to "reuse" $l$ instead of using $m$ (say) in $$ M_{jk}=\delta_{jk}+\epsilon_{jkl}n_l $$ when you already used it in $N_{ij}$.
Moral of the story: don't ever reuse a symbol for suffix, or if you must (because you ran out of letters), add some decorations like primes, tildes or subscripts of suffix to distinguish them.
Edit
So you want to evaluate $$ N_{ij}M_{jk}=(\delta_{ij}-\epsilon_{ijl}n_l+n_in_j)(\delta_{jk}+\epsilon_{jkm}n_m) $$ Expanding the RHS, $$\require{cancel} \begin{align*} RHS&=(\delta_{ij}-\epsilon_{ijl}n_l+n_in_j)\delta_{jk}+(\delta_{ij}-\epsilon_{ijl}n_l+n_in_j)\epsilon_{jkm}n_m\\ &=\delta_{ij}\delta_{jk}-\epsilon_{ijl}n_l\delta_{jk}+n_in_j\delta_{jk}+\delta_{ij}\epsilon_{jkm}n_m-\epsilon_{ijl}n_l\epsilon_{jkm}n_m+n_in_j\epsilon_{jkm}n_m\\ &=\delta_{ik}-\cancel{\epsilon_{ikl}n_l}+n_in_k+\cancel{\epsilon_{ikm}n_m}-\epsilon_{ijl}n_l\epsilon_{jkm}n_m+n_in_j\epsilon_{jkm}n_m \end{align*} $$ Note that $\epsilon_{jkm}$ is antisymmetric in $j,m$ but $n_mn_in_j$ is symmetric in $j,m$, so their contraction is gives 0. Also, recall the contraction of two $\epsilon$'s: $$ \epsilon_{ijl}\epsilon_{jkm}=\epsilon_{jli}\epsilon_{jkm}=\delta_{lk}\delta_{im}-\delta_{lm}\delta_{ik} $$ So this gives $$ \begin{align*} RHS&=\delta_{ik}+n_in_k-(\delta_{lk}\delta_{im}-\delta_{lm}\delta_{ik})n_ln_m\\ &=\delta_{ik}+\cancel{n_in_k}-\cancel{n_kn_i}+\delta_{ik}\underbrace{n_mn_m}_{=\mathbf{n}\cdot\mathbf{n}=1}\\ &=2\delta_{ik} \end{align*} $$
Of course, without using summation convention, you could approach it as: $N$ represents $\mathbf{x}\mapsto\mathbf{x}-\mathbf{x}\times\mathbf{n}+\mathbf{n}(\mathbf{n}\cdot\mathbf{x})$ and $M$ represents $\mathbf{x}\mapsto\mathbf{x}+\mathbf{x}\times\mathbf{n}$, so the composition is $$ \begin{align*} \mathbf{x}&\mapsto(\mathbf{x}+\mathbf{x}\times\mathbf{n})-(\mathbf{x}+\mathbf{x}\times\mathbf{n})\times\mathbf{n}+\mathbf{n}(\mathbf{n}\cdot(\mathbf{x}+\mathbf{x}\times\mathbf{n}))\\ &=(\mathbf{x}+\mathbf{x}\times\mathbf{n})-(\mathbf{x}\times\mathbf{n}-(\mathbf{n}\cdot\mathbf{n})\mathbf{x}+(\mathbf{x}\cdot\mathbf{n})\mathbf{n})+\mathbf{n}(\mathbf{n}\cdot\mathbf{x})\\ &=2\mathbf{x} \end{align*} $$