determinant of anti-diagonal matrix

Let $A=\mathrm{adiag}(a_1,\mathrm{\dots },a_n)$ be an anti-diagonal matrix. Using the sum over all permutations formula for the determinant of a matrix and since all but possibly the anti-diagonal elements are null we get directly at the result

$$\det A=\mathrm{sgn}(n,n-1,\mathrm{\dots },1)\prod _{i=1}^n{a}_i$$

so all that remains is to calculate the sign of the permutation.This can be done directly.

To bring the last element to the beginning $n-1$ permutations are needed so

$$\mathrm{sgn}(n,n-1,\mathrm{\dots },1)={(-1)}^{n-1}\mathrm{sgn}(1,n,n-1,\mathrm{\cdots },2)$$

Now bring the last element to the second position.To do this $n-2$ permutations are needed.Repeat this procedure $n-1$ times to get the permutation $(1,\mathrm{\dots },n)$ which has positive sign.

Summing every permutation, it takes

$$\sum _{k=1}^{n-1}k=\frac{n(n-1)}{2}$$

permutations to get to the desired permutation.

So we get the final result that

$$\det \mathrm{adiag}(a_1,\mathrm{\dots },a_n)={(-1)}^{\frac{n(n-1)}{2}}\prod _{i=1}^n{a}_i$$

Notice that the sign is positive if either $n$ or $n-1$ is a multiple of $4$ and negative otherwise.