So I'm reading this old paper by Cauchy's from 1847, and at one point, he merely states without proof the following identity: let $n$ be an odd prime number, and let $\rho$ be an $n^{\text{th}}$ root of unity. Then $$\prod_{k=1}^{n-1} (1-\rho^{k}) = n .$$ Due to the fact that Cauchy states it without proof and without reference to where proof can be found, I can only assume that it mustn't be too difficult. Nevertheless, I am unable to prove it on my own, and if it is a standard identity (and it certainly looks familiar), I have been unable to find it by googling it.
Any and all help would be appreciated.
Actually, this is a consequence of the fundamental theorem of algebra. Indeed, the $n^\mathrm{th}$ roots of unity, namely $z_k = \rho^k$, $k = 0,1,2,\ldots,n-1$, where $\rho = e^{2\pi i/n}$, are defined as the roots of the polynomial $z^n - 1$, which can be factorized as $$ z^n - 1 = \prod_{k=0}^{n-1} (z-z_k) = (z-1) \prod_{k=1}^{n-1} (z-\rho^k) $$ thanks to the said theorem, hence $$ \prod_{k=1}^{n-1} (z-\rho^k) = \frac{z^n-1}{z-1} $$ and finally $$ \prod_{k=1}^{n-1} (1-\rho^k) = n $$ by evaluating the expression at $z = 1$ with the help of L'Hospital's rule.