Edit: I know screen readers can't read images but I think this image adds more context regarding my notation which @Ian clarifies in the comments.
This is from the Graduate Algorithms course on Udacity with Georgia Tech using FFT for polynomial multiplication.
First row is all 1's. Second row is $1, W_n^{-1}$, ... $W_n^{-(n-1)}$ that can be rewritten not using negative powers as $1, W_n^{n-1}, W_n^{n-2},...,W_n^1$ The third row I can start but not finish: $1, W_n^{n-2}, W_n^{n-4},...,$
The last term in the third row with negative exponents is $W_n^{-2(n-1)}$ so I tried to use my limited algebra skills in figuring out what multiplied by $W_n^{2(n-1)}$ would equal $W_n^n = W_n^0$?
I let $2(n-1) + x = n$ and got $x = -n +2$ so I have for that last term in the third row, $W_n^{-2n+2}$ which can be rewritten as $W_n^{-2(n+1)}$ but I cannot visualize where this Omega would be on the unit circle on the complex plane where as for the second row I can clearly see that it starts at the last Omega and goes around clock-wise.
I followed @Ian's advice and tried for $n=8$
The last element of row 3 is $W_8^{2(8-1)}$ which is $W_8^{14}$ which on the unit circle on the complex plane is at $-i$ or in polar coordinates $(1, 6pi/4)$. Multiply this by $(1, 8pi/4)$ since $W_8^8$ is at 1 or $(1, 2pi)$ which gives me $(1, 14pi/4)$ which is at positive $i$. So the final term of the row 3 can be rewritten as $W_n^2$.