Result of conjugate symmetry property of DFT
The textbook probably even has the proof; but it's not hard to do: Simply write down the sum that gives you the $k$th bin of the transforf $X[k]$ of the time series $x[n]$. Then, flip $k$'s sign and apply the conjugate to finish your proof.

To show:

\begin{align}
X[k] &amp;\overset{\text{claim}}= X^*[-k]\\
X[k] &amp;= \sum_{n=0}^{N-1}x[n]e^{-j2\pi k\frac nN}\\
X^*[-k] &amp;= \left(\sum_{n=0}^{N-1}x[n]e^{-j2\pi (-k)\frac nN}\right)^*\\
&amp;= \sum_{n=0}^{N-1}\left(x[n]e^{-j2\pi (-k)\frac nN}\right)^*&amp;\text{Sum operator linear}\\
&amp;= \sum_{n=0}^{N-1}x[n]\left(e^{-j2\pi (-k)\frac nN}\right)^*&amp;x[n]=x^*[n]\text{ since real}\\
&amp;= \sum_{n=0}^{N-1}x[n]e^{-(-j2\pi (-k)\frac nN)}\\
&amp;= \sum_{n=0}^{N-1}x[n]e^{-j2\pi k\frac nN}\\
&amp;=X[k] &amp;\rule{1ex}{1ex}
\end{align}

Now, having showed that half of the coefficients are just the conjugate of the others: It's obviously sufficient to only consider one half; the other can be directly calculated from that.