To the power 99

Using Vieta's formulas , we have:

$\begin{aligned} \prod_{n=1}^{100} (x-(2n-1)) & = x^{100} - x^{99} \sum_{k=1}^{100} (2k-1) + x^{98} \sum_{j,k=1, j \ne k}^{100} (2j-1)(2k-1) + ... + \prod_{n=1}^{100} (2n-1) \end{aligned}$

The coefficient of $x^{99}$ is $-1\times$ sum of roots or $\begin{aligned} - \sum_{k=1}^{100} (2k-1) & = - \frac{100}2 (1+199) = \boxed {-10000} \end{aligned}$

There are $100$ terms in the expansion so the $x^{99}$ terms are formed by the $x's$ from $99$ brackets and the number from $1$ bracket. This gives the answer as:

$\sum_{i=1}^{100} -(2i-1)=- \left ( 2 \sum_{i=1}^{100} i - \sum_{i=1}^{100} 1 \right)=- \left ( 2 \times \dfrac{(100)(100+1)}{2}-100 \right)=\boxed{\boxed{-10000}}$