Factorizing a difference of two $n$-th powers How can be proved that
$$a^n-b^n=\displaystyle\prod_{j=1}^{n}(a-\omega^j b)$$
where $\omega=e^{\frac{2\pi i}{n}}$ is a primitive $n$-th root of $1$?

Fix $b$.
Let $$P(X):=X^n-b^n-\displaystyle\prod_{j=1}^{n}(X-\omega^j b) \,.$$
Then $P(X)$ is a polynomial of degree $n-1$ and $P(\omega^j b)=0$ for all $1 \leq j \leq n$.
Thus $P(X)=0$ for all $X$, and in particular $P(a)=0$.
