Problem:
Solution:
The hint basically spell out the solution. If the determinant is 0, then the columns are linearly dependent.
We can write, with $ c_i $ not all zero.
$ c_0\left(\begin{array}{c}1\\ \vdots \\ 1\end{array}\right) + c_1 \left(\begin{array}{c}a_1\\ \vdots \\ a_n\end{array}\right) + \cdots + c_n\left(\begin{array}{c}a_1^{n-1}\\ \vdots \\ a_n^{n-1}\end{array}\right) = 0 $
These equations can be interpreted as $ n $ roots of the polynomial $ c_0 + c_1 x + \cdots + c_n x^{n-1} $.
A non-zero polynomial of degree $ n -1 $ cannot have $ n $ roots, so we have a contradiction, the determinant is not 0.
Solution:
The hint basically spell out the solution. If the determinant is 0, then the columns are linearly dependent.
We can write, with $ c_i $ not all zero.
$ c_0\left(\begin{array}{c}1\\ \vdots \\ 1\end{array}\right) + c_1 \left(\begin{array}{c}a_1\\ \vdots \\ a_n\end{array}\right) + \cdots + c_n\left(\begin{array}{c}a_1^{n-1}\\ \vdots \\ a_n^{n-1}\end{array}\right) = 0 $
These equations can be interpreted as $ n $ roots of the polynomial $ c_0 + c_1 x + \cdots + c_n x^{n-1} $.
A non-zero polynomial of degree $ n -1 $ cannot have $ n $ roots, so we have a contradiction, the determinant is not 0.
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