Determinant of a cyclic permutation matrix

Permutation. $P$ sends $e_i$ to $e_{i+1}$ for $i<n$ and $e_n$ to $e_1$. It corresponds to the cycle $(1\,2\,\cdots\,n)$, which decomposes into $n-1$ transpositions.

Determinant. The determinant of a permutation matrix equals the sign of the permutation, so
$$\det P = (-1)^{n-1}.$$

Check. For $n=2$: $P=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $\det=-1=(-1)^1$. For $n=3$: $P=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$, $\det=1=(-1)^2$. Consistent.

Conclusion. $\det P = (-1)^{n-1}$.