Show $\prod \limits_{\substack{k=0\\k\neq k_0}}^{K-1}\left(\exp\left[{\frac{2 \pi i (k-k_0)}{K}}\right]-1\right)=(-1)^{K-1} K$ I know it is true that, for any $k_0 \text{ s.t. } 0\leq k_0 < K$,
$$\prod \limits_{\substack{k=0\\k\neq k_0}}^{K-1} \left(\exp\left[{\frac{2 \pi i (k-k_0)}{K}}\right]-1\right)=(-1)^{K-1} K$$
However, I'm unable to come up with a nice (hopefully intuitive) way to prove it.
Any suggestions?
Note: I know this is true because this is the expression for a residue of $z^{K-1}/(1+z^K)$, and I can compute the residue using limits.
(Thanks to David for corrections)

Unless I am missing something, your evaluation is not correct: for example, if $K=2$ and $k_0=1$ then your answer is $1$ but the correct product is $-2$.
We can evaluate the product as follows:
$$\eqalign{P
  &=\prod_{k=0}^{k_0-1}\left(\exp\left[{\frac{2 \pi i (k-k_0)}{K}}\right]-1\right)
    \prod_{k=k_0+1}^{K-1}
      \left(\exp\left[{\frac{2 \pi i (k-k_0)}{K}}\right]-1\right)\cr
  &=\prod_{k=K}^{K+k_0-1}
      \left(\exp\left[{\frac{2 \pi i (k-k_0)}{K}}\right]-1\right)    \prod_{k=k_0+1}^{K-1}
      \left(\exp\left[{\frac{2 \pi i (k-k_0)}{K}}\right]-1\right)\cr
    &=\prod_{k=k_0+1}^{K+k_0-1}\left(\exp\left[{\frac{2 \pi i (k-k_0)}{K}}\right]-1\right)\cr
  &=\prod_{m=1}^{K-1}
    \left(\exp\left[{\frac{2 \pi im}{K}}\right]-1\right)\ .\cr}$$
If we write
$$f(x)=\prod_{m=1}^{K-1}
    \left(x-\exp\left[{\frac{2 \pi im}{K}}\right]\right)=x^{K-1}+x^{K-2}+\cdots+x+1$$
then
$$P=(-1)^{K-1}f(1)=(-1)^{K-1}K\ .$$
Addendum.  Formula for $f(x)$: the roots of $x^K=1$ are $x=\exp(2\pi i m/K)$ for $m=0,1,\ldots,K-1$.  Therefore
$$x^K-1=\prod_{m=0}^{K-1}
    \left(x-\exp\left[{\frac{2 \pi im}{K}}\right]\right)\ ,$$
and dividing both sides by $x-1$ gives
$$x^{K-1}+x^{K-2}+\cdots+x+1
 =\prod_{m=1}^{K-1}\left(x-\exp\left[{\frac{2 \pi im}{K}}\right]\right)\ .$$
