Prove this product: $\prod\limits_{k=2}^ n {\frac{k^2+k+1}{k^2-k+1}}=\frac{n^2+n+1}{3}$ How to prove this product? 
$$\prod\limits_{k=2}^ n {\frac{k^2+k+1}{k^2-k+1}}=\frac{n^2+n+1}{3}$$

$$\require{cancel}\prod_{k=2}^n\frac{k^2+k+1}{k^2-k+1}=\frac{\cancel7}{3}\frac{\cancel{13}}{\cancel7}\frac{\cancel{21}}{\cancel{13}}\frac{31}{\cancel{21}}\cdot\ldots$$
Can you see what the only factors that'll remain are?
