Why is the second half of the FFT negative frequencies
If you say that plugging in a negative value for $k$ would make sense to you then have a look at this:

$$\begin{align}X[N-1]&amp;=\sum_{n=0}^{N-1}x[n]e^{-j2\pi (N-1)n/N}\\&amp;=\sum_{n=0}^{N-1}x[n]e^{-j2\pi n}e^{j2\pi n/N}\\&amp;=\sum_{n=0}^{N-1}x[n]e^{j2\pi n/N}=X[-1]\end{align}$$

The DFT $X[k]$ is periodic with period $N$, so it doesn't matter if you consider the interval $k\in[0,N-1]$, or $k\in[-N/2+1,N/2]$ (assuming $N$ is even).