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For $ngeq 2$, $x_1,x_2,..,x_n$ distinct real numbers and $alpha > 0$,$alpha neq 1$ a real number, show that the determinant of the matix

beginbmatrix
1 & 1 & 1 & ... & 1 \
x_1 & x_2 & x_3 & ... & x_n \
vdots & vdots &vdots & ddots & vdots \
x_1^n-2 & x_2^n-2& x_3^n-2 & ... & x_n^n-2 \
alpha^x_1 & alpha^x_2 & alpha^x_3 & ... & alpha^x_n
endbmatrix

is nonzero.

My attempts:

For $n = 2$ we get $alpha^x_1 - alpha^x_2 neq 0$ which is clear.

For n = 3 we get $alpha^x(z - y) + alpha^y(x - z) +alpha^z(y - x) neq 0$ (I have renamed $x_1,x_2,x_3$ to $x,y,z$ for simplicity). If you move the middle term to the other side and divide by $alpha^y$ and set $a = x-y, b = z-y$ this turns into
$$fracalpha^a - 1a neq fracalpha^b - 1b.$$
For $alpha$ different from 1, it is clear that the function $fracalpha^x - 1x$ is strictly increasing or strictly decreasing, depending on $alpha$. Thus it is injective, and so $fracalpha^a - 1a = fracalpha^b - 1b iff a = b iff x = z$, which is a contradiction.

Now for $n geq 4$.

If you expand the determinant by the second-to-last line, you get a linear combination of the $x_i^n-2$, in which the sum of the coefficients is zero (you can see that by replacing the second-to-last line with a line full of 1's; The determinant would be zero since the first line would be equal to the second-to-last line, and by expanding the determinant you get that the sum of coefficients is zero). Since by induction the coefficients are nonzero, it would amount to showing that if a linear combination of the $x_i^n-2$'s with coefficients summing to zero is zero, then at least one coefficient is zero, which does not seem right after some experimenting.

The other way is to expand it by the last line; I've explored that in this post: $a_1 k^b_1 + a_2 k^b_2 + ... + a_n k^b_n = 0$ I have not been able to prove that rigorously.

Any ideas would be appreciated.