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? 
$$\prod\limits_{k=2}^ n {\frac{k^2+k+1}{k^2-k+1}}=\frac{n^2+n+1}{3}$$
 A: $$\require{cancel}\prod_{k=2}^n\frac{k^2+k+1}{k^2-k+1}=\frac{\cancel7}{3}\frac{\cancel{13}}{\cancel7}\frac{\cancel{21}}{\cancel{13}}\frac{31}{\cancel{21}}\cdot\ldots$$
Can you see what the only factors that'll remain are?
A: 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. 
A: HINT: It never hurts to gather some data by doing some actual computation:
$$\begin{array}{c|l}
n&\prod_{k=2}^n\frac{k^2+k+1}{k^2-k+1}\\ \hline
2&\frac73\\
3&\frac{\color{red}7}3\cdot\frac{13}{\color{red}7}\\
4&\frac{\color{red}7}3\cdot\frac{\color{blue}{13}}{\color{red}7}\cdot\frac{21}{\color{blue}{13}}\\
5&\frac{\color{red}7}3\cdot\frac{\color{blue}{13}}{\color{red}7}\cdot\frac{\color{green}{21}}{\color{blue}{13}}\cdot\frac{31}{\color{green}{21}}
\end{array}$$
A: Note that $$k^{2}+k+1=\left(k+e^{i\pi/3}\right)\left(k-e^{i\pi/3}+1\right)
 $$ and $$k^{2}-k+1=\left(k-e^{i\pi/3}\right)\left(k+e^{i\pi/3}-1\right)
 $$ where $e^{i\pi/3}$ is the third roots of the unity. So $$\prod_{k=2}^{n}\frac{k^{2}+k+1}{k^{2}-k+1}=\prod_{k=2}^{n}\frac{\left(k+e^{i\pi/3}\right)\left(k-e^{i\pi/3}+1\right)}{\left(k-e^{i\pi/3}\right)\left(k+e^{i\pi/3}-1\right)}
 $$ $$=\prod_{k=2}^{n}\frac{k+e^{i\pi/3}}{k-e^{i\pi/3}}\prod_{k=3}^{n+1}k-e^{i\pi/3}\prod_{k=1}^{n-1}\frac{1}{k+e^{i\pi/3}}
 $$ $$=\frac{\left(n+e^{i\pi/3}\right)\left(n+1-e^{i\pi/3}\right)}{\left(1+e^{i\pi/3}\right)\left(2-e^{i\pi/3}\right)}=\color{red}{\frac{n^{2}+n+1}{3}}.$$
A: Once you start writing all terms, you will notice that only the numerator of last k, which is $ n^2 + n +1 $ and denominator of first term, which is 3, will be left after cancellation
