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

Hint: If $f(m)=m^2+m+1$ then $f(m-1)=m^2-m+1$. Now there will be a whole lot of cancellation goin' on.

HINT: It never hurts to gather some data by doing some actual computation:

$$\begin{array}{c|l} n&\prod_{k=2}^n\frac{k^2+k+1}{k^2-k+1}\\ \hline 2&\frac73\\ 3&\frac{\color{red}7}3\cdot\frac{13}{\color{red}7}\\ 4&\frac{\color{red}7}3\cdot\frac{\color{blue}{13}}{\color{red}7}\cdot\frac{21}{\color{blue}{13}}\\ 5&\frac{\color{red}7}3\cdot\frac{\color{blue}{13}}{\color{red}7}\cdot\frac{\color{green}{21}}{\color{blue}{13}}\cdot\frac{31}{\color{green}{21}} \end{array}$$

$$\require{cancel}\prod_{k=2}^n\frac{k^2+k+1}{k^2-k+1}=\frac{\cancel7}{3}\frac{\cancel{13}}{\cancel7}\frac{\cancel{21}}{\cancel{13}}\frac{31}{\cancel{21}}\cdot\ldots$$

Can you see what the only factors that'll remain are?